10th_class_maths_textbook.pdf to read 10th class students

INSTRUCTIONS TO STUDENTS
F Pictures/figures are provided for many items/situations for better understanding of
the concepts in the text book.You have to read them carefully in relation with the
concepts.
F Clarify your doubts with the help of your teachers while you are doing the activites
related to the concepts time to time.
F ''Do this”Problems are meant to test your progress immediately after learning a
concept.So you have to do these problems on your own.Check these with your teacher.
F ”Try these”items in every chapter are provided to sharpen your ideas and
thoughts.some logic will be there behind these questions.You have to try all these
questions with out fail.
F Do this and Try these problems are meant to solve in the class room under the
guidance of your teacher.Do not leave them at all.
F The questions under “Think discuss and write” are given to test your analytical ability.You
can answer these questions by discussing in groups.
F The exercises given at the end of some sub topics arerelated to the concepts learned
to a particular level.You try to solve these problems on your own.Do not copy them
from guides and from other sources.
F Some projects given in in the textbook are meant to be done with proper plan and
guidance with the teacher.you can do this in a group.A report is to be submitted at the
end of the project.
F Some blank spaces are provided in tables,try these exercises to facilitate you to
answer quickely.You can fill it in the text book it self.
F Every student is required to finish the day's work on the same day it self.Donot post
pone it.You may feel burden later.
F Try to collect some more information to create some similar problems based on the
concepts you have learned.Ask your friends to solve your questions.
F Games,Puzzles provided in the textbooks shall encourage you to feel joy in doing
maths.For recreation collect some more .
F You should always connect the concepts learned in the class with other subject ar-
eas and situations out side the class room.
F You have to develop abilities with regard to academic standards prescribed at the
end of the text book based on the syllabus .The academic standards at the second-
ary level are based on a) Problem solving 2)Reasoning and Proof 3)Mathematical
communication 4)Connections 5)Representation.
F You have to go through various reference books,Internet,Forums for finding different
ways of solutions for a problem and for better understanding of the concepts.
F Optional exercises are provided to think critically to solve certain problems.These
exercises will help you to face Talent tests conducted at national level on par with
other students.
F The appendix chapter “Mathematical modelling” will help you to know modelling of
mathematical concepts in the fields of business,Agriculture,Stock markets etc,You
can see the connection of mathematics with different areas.
F Syllabus is also provided at the end of the textbook for your reference that how you
are making your progress.You can also compare with others.
We wish you all the best
SIGNSANDSYMBOLSOFSCHOOLMATHEMATICS
Sign/symbol Read as Mathematical meaning
± plus or minus Add or Substract
≠ not equal to unequal
∴ therefore logical flow of a statement
∞ infinite not finite
∼ is similar to same in geometrical shape
≅ is congruent to same shape and same size
≡ is identically equal to equivalent statements
∀ for all universal quantifier
square root of square root of a number
3
cube root of cube root of a number
∪ cup of union of sets
∩ cap of intersection of sets
φ phi symbol for golden ratio
% percent of per hundred
o degree angle measure
Δ delta / triangle symmetric difference in sets/symbol of triangle
∈ belongs to an element belong to a particular set
↔ equivilent to one to one correspondence
α, β, γ alfa,beta,gamma greek lettetrs to represent zeroes of polynomial
μ mu universal set symbol
π pi circumference of a circle / diameter
σ sigma sum of scores
sin θ, cos θ, tan θ sin theta,cos theta,tan theta trigonometric ratios
x x bar arithmetic mean
loga
x log x to the base a logarthemic function
(a, b) point (a, b) ordered pair (a, b)
|x| mod x absolute value of a real number
P(x) P of x a polynomial function in x
P(E) P of E probability of an event
∵ Since reasoning at a stage
` rupee symbol of Indian rupee
| | is parallel to parallel lines
⊥ is perpendicular to making 90 degree with a line
{ } flower bracket used to set notation
PQ arc PQ arc of a circle
a2
a square square of a number
∠ angle symbol of angle
θ theta measurement of an angle

Textbook Development & Publishing Committee
ChiefProductionOfficer : Sri. G. Gopal Reddy,
Director, SCERT, Hyderabad.
Executive ChiefOrganiser : Sri. B. Sudhakar,
Director, Govt. Text Book Press, Hyderabad.
OrganisingIncharge : Dr. Nannuru UpenderReddy,
Prof. & Head, Curriculum&Text Book Department,
SCERT, Hyderabad.
CHAIRPERSON FOR POSITION PAPER AND MATHEMATICS CURRICULUM AND TEXTBOOK
DEVELOPMENT
Prof.V.Kannan
Department of Mathematics and Statistics, HCU, Hyderabad
CHIEF ADVISORS
Sri Chukka Ramaiah Dr. H.K.Dewan
Eminent Scholar in Mathematics, Educational Advisor, Vidya Bhavan Society
Telangana, Hyderabad. Udaipur, Rajasthan
Published by:
The Government of Telangana, Hyderabad
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Mathematics Class-X
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© Government of Telangana, Hyderabad.
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WRITERS
RITERS
RITERS
RITERS
RITERS
Sri TataVenkata Rama Kumar Sri Gottumukkala V.B.S.N. Raju
H.M., ZPHS, Mulumudi, SPS Nellore SA, Mpl. High School, Kaspa, Vizianagaram.
Sri Soma Prasada Babu Sri Padala Suresh Kumar
PGT. APTWRS, Chandrashekarapuram, SPS Nellore SA,GHS, Vijayanagar Colony, Hyderabad.
Sri G.Anantha Reddy Sri PeddadaD.L.GanapatiSharma
Retd. Headmaster, Ranga Reddy SA,GHS, Zamisthanpur, Manikeshwar Nagar, Hyd.
Dr. Poondla Ramesh Sri Sardar Dharmendra Singh
Lecturer, Government lASE, SPS Nellore SA, ZPHS, Dannur (B), Adilabad
Sri Komanduru Sreedharacharyulu Sri Nagula Ravi
SA, ZPHS, Rangaipally, Medak SA, ZPHS, Lokeshwaram, Adilabad
Sri KandalaRamaiah Sri KakulavaramRajenderReddy
SA, ZPHS, Kasimdevpet, Warangal Co-ordinator, SCERT, T.S., Hyderabad.
Sri Ramadugu Lakshmi Narsimha Murthy
SA, ZPHS, Thupranpet, Nalgonda
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Dr. H.K.Dewan
Educational Advisor, Vidya Bhavan Society
Udaipur, Rajasthan
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EDITORS
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DITORS
Prof. V. Shiva Ramaprasad Prof. N.Ch.Pattabhi Ramacharyulu
Retd. Dept. of Mathematics, Retd., National Institute of Technology,
Osmania University, Hyderabad Warangal
Sri A. Padmanabham (Rtd.) Dr. G.S.N. Murthy (Rtd.)
Head of the Dept. of Mathematics Reader in Mathematics
Maharanee College, Rajah R.S.R.K.Ranga Rao College,
Peddapuram, East Godavari Dist. Bobbili, Vizianagaram Dist. (A.P.)
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Sri Kakulavaram Rajender Reddy Sri K. Narayan Reddy
Co-ordinator, SCERT, Hyderabad. Lecturer, SCERT, Hyderabad
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Sri Hanif Paliwal Ms. Preeti Mishra
Vidya Bhawan Education Resource Centre, Udaipur Vidya Bhawan Education Resource Centre, Udaipur
Mrs. Snehbala Joshi Ms.Tanya Saxena
Ms. M.Archana
Department of Mathematics and Statistics, University of Hyderabad
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Sri. Prashant Soni
Illustrator, Vidya Bhawan Education Resource Centre, Udaipur
Sri S. M. Ikram Sri Bhawani Shanker
DTP Operator, DTP Operator,
Sri Sunkara Koteswara Rao Smt. Sunkara Sunitha
DTP Operator, DTP Operator,
Pavan Graphics, Vignanpuricolony,Vidyanagar, Hyderabad. Pavan Graphics, Vignanpuricolony,Vidyanagar, Hyderabad.
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Sri. K. Sudhakara Chary, HM, UPS Neelikurthy, Mdl.Maripeda, Dist. Warangal

Foreword
Education is a process of human enlightenment and empowerment. Recognizing
the enormous potential of education, all progressive societies have committed themselves
to the Universalization of Elementary Education with a strong determination to provide
quality education to all. As a part of its continuation, universalization of Secondary
Education has gained momentum.
In the secondary stage, the beginning of the transition from functional mathematics
studied upto the primary stage to the study of mathematics as a discipline takes place.
The logical proofs of propositions, theorems etc. are introduced at this stage. Apart from
being a specific subject, it is connected to other subjects involving analysis and through
concomitant methods. It is important that children finish the secondary level with the
sense of confidence to use mathematics in organising experience and motivation to
continue learning in High level and become good citizens of India.
I am confident that the children in our state Telangana learn to enjoy mathematics,
make mathematics a part of their life experience, pose and solve meaningful problems,
understand the basic structure of mathematics by reading this text book.
For teachers, to understand and absorb critical issues on curricular and pedagogic
perspectives duly focusing on learning in place of marks, is the need of the hour. Also
coping with a mixed class room environment is essentially required for effective transaction
of curriculum in teaching learning process. Nurturing class room culture to inculcate
positive interest among children with difference in opinions and presumptions of life style,
to infuse life in to knowledge is a thrust in the teaching job.
The afore said vision of mathematics teaching presented in State Curriculum Frame
work (SCF -2011) has been elaborated in its mathematics position paper which also
clearly lays down the academic standards of mathematics teaching in the state. The text
books make an attempt to concretize all the sentiments.
The State Council for Education Research and Training Telangana appreciates the
hard work of the text book development committee and several teachers from all over the
state who have contributed to the development of this text book. I am thankful to the
District Educational Officers, Mandal Educational Officers and head teachers for making
this possible. I also thank the institutions and organizations which have given their time in
the development of this text book. I am grateful to the office of the Commissioner and
Director of School Education for extending co-operation in developing this text book. In
the endeavor to continuously improve the quality of our work, we welcome your comments
and suggestions in this regard.
Place : Hyderabad Director
Date : 17 October, 2013 SCERT, Hyderabad
(iv)

With this Mathematics book, children would have completed the three years of
learning in the elimentary classes and one year of secondary class. We hope that
Mathematics learning continues for all children in class X also however, there may be
some children from whom this would be the last year of school. It is, therefore, important
that children finish the secondary level with a sense of confidence to use Mathematics in
organizing experience and motivation to continue learning.
Mathematics is essential for everyone and is a part of the compulsory program for
school education till the secondary stage. However, at present, Mathematics learning does
not instill a feeling of comfort and confidence in children and adults. It is considered to be
extremely difficult and only for a few. The fear of Mathematics pervades not just children
and teachers but our entire society. In a context where Mathematics is an increasing part
of our lives and is important for furthering our learning, this fear has to be removed. The
effort in school should be to empower children and make them feel capable of learning
and doing Mathematics. They should not only be comfortable with the Mathematics in the
classroom but should be able to use it in the wider world by relating concepts and ideas of
Mathematics to formulate their understanding of the world.
One of the challenges that Mathematics teaching faces is in the way it is defined.
The visualization of Mathematics remains centered around numbers, complicated
calculations, algorithms, definitions and memorization of facts, short-cuts and solutions
including proofs. Engaging with exploration and new thoughts is discouraged as the
common belief is that there can be only one correct way to solve a problem and that
Mathematics does not have possibilities of multiple solutions.
Through this book we want to emphasize the need for multiple ways of attempting
problems, understanding that Mathematics is about exploring patterns, discovering
relationships and building logic. We would like all teachers to engage students in reading
the book and help them in formulating and articulating their understanding of different
concepts as well as finding a variety of solutions for each problem. The emphasis in this
book is also on allowing children to work with each other in groups and make an attempt
to solve problems collectively. We want them to talk to each other about Mathematics and
create problems based on the concepts that have learnt. We want everybody to recognize
that Mathematics is not only about solving problems set by others or learning proofs and
methods that are developed by others, but is about exploration and building new arguments.
Doing and learning Mathematics is therefore about each person coming up with her own
methods and own rules.
Preface
(v)

Class X is the final year of secondary level students and their have already dealt
about the consolidation of initiations. They have already learnt also to understand that
Mathematics consists of ideas that are applied in life situations but do not necessarily
arise from life. We would also like children to be exposedto the notion of proof and recognize
that presenting examples is not equivalent to proof with modeling aspects.
The purpose of Mathematics as we have tried to indicate in the preface as well as
in the book has widened to include exploring mathematization of experiences. This means
that students can begin to relate the seemingly abstract ideas they learn in the classrooms
to their own experiences and organize their experiences using these ideas. This requires
them to have opportunity to reflect and express both their new formulations as well as
their hesitant attempt on mathematizing events around them. We have always emphasized
the importance of language and Mathematics interplay. While we have tried to indicate at
many places the opportunity that has to be provided to children to reflect and use language.
We would emphasise the need to make more of this possible in the classrooms. We
have also tried to keep the language simple and close to the language that the child normally
uses. We hope that teachers and those who formulate assessment tasks would recognize
the spirit of the book. The book has been developed with wide consultations and I must
thank all those who have contributed to its development. The group of authors drawn from
different experiences have worked really hard and together as a team. I salute each of
them and look forward to comments and suggestions of those who would be users of this
book.
Text Book Development Committee
(vi)

Mathematics Class-X
(vii)
CHAPTER CONTENTS NO. OF SYLLABUS TO BE PAGE
NUMBER PERIODS COVERED DURING NUMBER
01 Real Numbers 15 June 1 - 28
02 Sets 08 June 29 - 50
03 Polynomials 08 July 51 - 76
04 Pair of Linear Equations in 15 September 77 - 104
Two Variables
05 Quadratic Equations 12 October 105 - 128
06 Progressions 11 January 129 - 162
07 Coordinate Geometry 12 November 163- 194
08 Similar Triangles 18 July,August 195 - 228
09 Tangents and Secants to a Circle 15 November 229 - 248
10 Mensuration 10 December 249 - 272
11 Trigonometry 15 August 273 - 297
12 Applications of Trigonometry 08 September 298 - 308
13 Probability 10 January 309 - 326
14 Statistics 15 July 327 - 356
Appendix Mathematical Modelling 08 January 357 - 369
Answers 370 - 388
Revision February

OUR NATIONAL ANTHEM
- Rabindranath Tagore
Jana-gana-mana-adhinayaka, jaya he
Bharata-bhagya-vidhata.
Punjab-Sindh-Gujarat-Maratha
Dravida-Utkala-Banga
Vindhya-Himachala-Yamuna-Ganga
Uchchala-Jaladhi-taranga.
Tava shubha name jage,
Tava shubha asisa mage,
Gahe tava jaya gatha,
Jana-gana-mangala-dayaka jaya he
Bharata-bhagya-vidhata.
Jaya he, jaya he, jaya he,
Jaya jaya jaya, jaya he!
“India is my country. All Indians are my brothers and sisters.
I love my country, and I am proud of its rich and varied heritage.
I shall always strive to be worthy of it.
I shall give my parents, teachers and all elders respect,
and treat everyone with courtesy. I shall be kind to animals
To my country and my people, I pledge my devotion.
In their well-being and prosperity alone lies my happiness.”
PLEDGE
(viii)
- Pydimarri Venkata Subba Rao

1.1 INTRODUCTION
"God made the integers. Allelse is the work ofman" - Leopold Kronecker.
Life is fullofnumbers. Imaginethe moment you wereborn. Your parentsprobablynoted
the time you were born, yourweight, your length and the most important, counted your fingers
and toes. Fromthen, numbers accompanyyou throughout life.
What are the other contexts where you dealwithnumbers?
Weusethenumbersto tellour agetokeeptrack ofour incomeandtofindthesavingsafter
spending certainmoney. We measure our wealth also.
In this chapter we are going to explore the notion of the numbers. Numbers play a
fundamental role within the realm of mathematics. We will come to see the richness of
numbers and delve into their surprising traits. Some collection of numbers fit so well
together that they actually lead to notions of aesthetics and beauty.
Let us look in to a puzzle.
In a garden of flowers, a swarm of bees is setting in equal number on flowers. When
they settle on two flowers, one bee will be left out. When they settle on three flowers, two
bees will be left out. When they settle on four flowers, three bees will be left out. Similarly,
when they settle on five flowers no bee will be left out. If there are at most fifty bees, how
many bees are there in the swarm?
Let us analyse and solve this puzzle.
Let the number of bees be 'x'. Then working backwards we see that x < 50.
If the swarmofbeesis divided into 5 equalgroups no bee willbe left, whichtranslates to
x = 5a + 0 for some natural number 'a'.
Real Numbers
1

Class-XMathematics
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FreeDistributionbyT.S.Government2019-20
If the swarm is divided in to 4 equal groups 3 bees will be left out and it translates to
x = 4b + 3 for some natural number b.
If the swarm is divided into 3 equal groups 2 bees will be left out and it translates to
x = 3c + 2 for some natural number c.
If the swarm is divided into 2 equal groups 1 bee will be left out and it translates to
x = 2d + 1 for some natural number d.
That is, in each case we have a positive integer y (in this example y takes values 5, 4, 3
and 2 respectively) which divides x and leaves remainder 'r' (in our case r is 0, 3, 2 and 1
respectively), that is smallerthan y. In the process ofwriting above equations, unknowingly,
we have usedDivision Algorithm.
Getting backto our puzzle.Tosolve it wemust look for themultiplesof5, whichsatisfyall
the conditions, because x = 5a + 0.
If a number leaves remainder 1 when it is divided by 2 we must consider odd multiples
only. In this case we have 5, 15, 25, 35 and 45. Similarly if we check for the remaining two
conditions youwillget 35.
Therefore, the swarm of bees contains 35 bees.
Let usverifythe answer.
When 35 is divided by2, the remainder is 1. That can be written as
35 = 2 ´ 17 + 1
35 = 3 ´ 11 + 2
35 = 4 ´ 8 + 3
and when35 is divided by5, the remainder is '0'. That can be written as
35 = 5 ´ 7 + 0
Let us generalise this. For each pair of positive integers a and b (dividend and divisor
respectively), we can find the whole numbers q and r (quotient and remainder respectively)
satifyingthe relation
a = bq + r, 0 < r < b

RealNumbers 3
DO THIS
Findqandrfor thefollowingpairsofpositiveintegersaandb, satisfyinga=bq+r.
(i) a = 13, b = 3 (ii) a = 80, b = 8 (iii) a = 125, b = 5
(iv) a = 132, b = 11
THINK AND DISCUSS
In questions of above "DO THIS", what is the nature of q and r?
Theorem-1.1 : (DivisionAlgorithm) : Given positive integers a and b, thereexist unique pair
of integers q and r satisfying a = bq + r, 0 < r < b.
This result, was first recorded in Book VII of Euclid's Elements. Euclid's algorithmis
based onthis division algorithm.
Euclid's algorithm is a technique to compute the Highest common factor (HCF) of two
given integers. Recall that the HCF of two positive integers a and b is the greatest positive
integer d that divides both a and b.
Let us find the HCF of60 and 100, throughthe following activity.
ACTIVITY
Take two paper strips of equalwidth and having lengths 60 cm, and 100 cmlong. Our
task is to find the maximum length of a strip which can measure both the strips without
leaving anypart.
Take 60cmstrip and measurethe 100 cmstripwith it. Cut offthe left over 40 cm. Now,
take this40 cmstrip andmeasure the 60 cmstrip with it. Cut offthe left over 20 cm. Now,
take this 20 cmstrip and measure the 40cmwith it.
Since nothingisleft over, wemayconclude that 20cmstrip isthe longest strip whichcan
measure both60 cmand 100 cmstrips without leaving anypart.

Class-XMathematics
4
Let uslink the process wefollwed inthe"Activity"to Euclid'salgorithmto get HCFof60
and 100.
When 100 is divided by 60, the remainder is 40
100 = (60 ´ 1) + 40
Now consider the division of 60 with the remainder 40 in the above equation and apply
thedivisionalgorithim
60 = (40 ´ 1) + 20
Now consider the division of 40 with the remainder 20, and applythe division lemma
40 = (20 ´ 2) + 0
Notice that the remainder has become zero and we cannot proceed anyfurther. We claim
that the HCFof60 and 100is thedivisor at this stage, i.e. 20. (Youcaneasilyverifythis bylisting
allthe factorsof60 and 100.)We observe that it is a series ofwell defined steps to find HCF of
60 and 100. So, let us state Euclid's algorithm clearly.
To obtain the HCFoftwo positive integers, say c and dwithc > d, followthesteps below:
Step 1 : ApplyEuclid's divisionlemma, to c and d. So,we find unique pair ofwhole numbers,
q and r such that c = dq + r, 0 < r < d.
Step 2 : If r = 0, d is the HCF of c and d. If r ¹ 0, apply the division lemma to d and r.
Step 3 : Continue the process tillthe remainder is zero. The divisor at this stage will be the
required HCF.
This algorithm works because HCF (c, d) = HCF (d, r) where the symbol HCF (m, n)
denotes the HCF ofanytwo positive integers m and n.
DO THIS
Find the HCF of the following by using Euclid algorithm.
(i) 50 and 70 (ii) 96 and 72 (iii) 300 and 550
(iv) 1860 and 2015

RealNumbers 5
THINK AND DISCUSS
Can you find the HCF of1.2 and 0.12? Justifyyour answer.
Euclid'salgorithmisusefulfor calculatingtheHCFofverylargenumbers,and it wasoneof
the earliest examplesofanalgorithmthat a computer had beenprogrammed to carryout.
Remarks :
1. Euclid's algorithmand divisionalgorithmareso closelyinterlinked that people oftencall
former as thedivisionalgorithmalso.
2. AlthoughDivisionAlgorithmisstated for onlypositive integers, it canbeextended forall
integers a and b where b ¹ 0. However, we shallnot discuss this aspect here.
Division algorithmhas severalapplications relatedto finding properties ofnumbers. We
give some examplesof these applications below:
Example 1 : Showthat everypositiveeveninteger isoftheform2q, andthat everypositiveodd
integer is ofthe form2q + 1, where q is some integer.
Solution : Let a be anypositive integer and b = 2. Then, bydivision algorithm, a = 2q + r, for
some integer q > 0, and r = 0 or r = 1, because 0 < r < 2. So, a = 2q or 2q + 1.
If a isofthe form2q, thena is an even integer. Also, a positive integer canbe either even
or odd. Therefore, anypositive odd integer is ofthe form2q + 1.
Example 2 : Show that everypositive odd integer is ofthe form4q + 1 or 4q + 3, where q is
someinteger.
Solution : Let a be a positive odd integer, and b = 4. We applythe divisionalgorithmfor aand
b = 4.
We get a = 4q +r, for q > 0, and 0 < r < 4. The possible remainders are 0, 1, 2 and 3.
That is, a canbe 4q, 4q + 1, 4q + 2, or 4q + 3, where q is the quotient. However, since
a is odd, a cannot be 4q = 2 (2q) or 4q + 2 = 2(2q+1) (since they are both divisible by 2).
Therefore, anyodd integer is of the form4q + 1 or 4q + 3.

Class-XMathematics
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EXERCISE - 1.1
1. Use Euclid'salgorithmto find the HCF of
(i) 900 and 270 (ii) 196 and 38220 (iii) 1651 and 2032
2. Usedivision algorithmto show that anypositive odd integer isofthe form6q+ 1,
or 6q + 3 or 6q + 5, where q is some integer.
3. Use divisionalgorithmto show that the square ofanypositive integer is ofthe form
3p or 3p + 1.
4. Use division algorithmto show that thecube ofanypositive integer is ofthe form9 m,
9m + 1 or 9m + 8.
5. Show that one and only one out of n, n + 2 or n + 4 is divisible by 3, where n is any
positive integer.
1.2 THE FUNDAMENTAL THEOREM OF ARITHMETIC
We know from DivisionAlgorithmthat for given positive integers a and b there exist
unique pair ofintegers q and r satifying
a = bq + r, 0 < r < b
THINK - DISCUSS
If r = 0, then what is the relationship between a, b and q in a = bq + r ?
Fromthe above discussion you might have concluded that if a = bq, 'a' isdivisible by
'b' then we can say that 'b' is a factor of 'a'.
For example we know that 24 = 2 ≥ 12
24 = 8 ≥ 3
= 2 ≥ 2 ≥ 2 ≥ 3
We know that, if 24 = 2 ≥12 then we can say that 2 and 12 are factors of 24. We can
also write 24 = 2 ≥ 2 ≥ 2 ≥ 3 and you know that this is the prime factorisation of 24.

RealNumbers 7
Let us take anycollection ofprime numbers, say2, 3, 7, 11 and 23. Ifwe multiplysome
or all of these numbers, allowing them to repeat as many times as we wish, we can produce
infinitelymanylarge positive integers. Let us observe afew :
2 ≥ 3 ≥ 11 = 66 7 ≥ 11 ≥ 23 = 1771
3 ≥ 7 ≥ 11
1 ≥ 23 = 5313 2 ≥ 3 ≥ 7 ≥ 11
1 ≥ 23 = 10626
23
≥ 3 ≥ 73
= 8232 22
≥ 3 ≥ 7 ≥ 11
1 ≥ 23 = 21252
Now, let us suppose your collection ofprimes includes allthe possible primes. What is
your guess about the size of this collection? Does it contain only a finite number of primes or
infinitelymany? Infact, there are infinitelymanyprimes. So, ifwe multiplyalltheseprimes in all
possible ways, wewillget an infinite collectionofcomposite numbers.
Now, let us consider the converseofthis statement i.e. ifwe take acomposite number
canit be written as aproduct ofprime numbers?The following theoremanswersthe question.
Theorem-1.2 : (FundamentalTheoremofArithmetic) : Every composite numbercan be
expressed (factorised) as a product of primes, and this factorization is unique, apart
from theorderin which the prime factors occur.
This givesus the FundamentalTheoremofArithmetic which says that every composite
number can be factorized as a product of primes.Actually, it says more. It says that anygiven
composite number can be factorized as a product ofprime numbers in a ‘unique’ way, except
for the order in which the primes occur. For example, when we factorize 210, we regard 2 ≥ 3
≥ 5 ≥ 7 as same as 3 ≥ 5 ≥ 7 ≥ 2, or any other possible order in which these primes are
written. That is, given any composite number there is one and only one way to write it as a
product of primes, as long as we are not particular about the order in whichthe primes occur.
In general, given a composite number x, we factorize it as x = p1
.p2
.p3
.....pn
, where
p1
, p2
, p3
...., pn
are primes and written in ascending order, i.e., p1 £ p2 £... £pn
. If we
combine the equal primes, we will get powers of primes. Once we have decided that the
order will be ascending, then the way the number is factorised, is unique. For example,
27300 = 2 ≥ 2 ≥ 3 ≥ 5 ≥ 5 ≥ 7 ≥13 = 22 ≥ 3 ≥ 52 ≥ 7 ≥ 13
DO THIS
Express 2310 as a product of prime factors. Also see how your friends have
factorized the number. Have they done it same as you? Verify your final product with
your friend’s result. Try this for 3 or 4 more numbers. What do you conclude?

Class-XMathematics
8
Let us applyfundamentaltheoremofarithmetic
Example 3. Consider the numbersofthe form4n
where n is a naturalnumber. Check whether
there is anyvalue of n for which 4n
ends with zero?
Solution : If4n
isto end with zero for a naturalnumber n, it should be divisible by2 and 5. This
means that the prime factorisation of4n
should contain the prime number 5 and 2. But it is not
possible because 4n
= (2)2n
so 2 isthe onlyprime in the factorisationof4n
. Since 5is not present
in the primefactorization, there is no naturalnumber n for which 4n
endswith the digit zero.
You have already learnt how to find the HCF (Highest Common Factor) and LCM
(Lowest Common Multiple) of two positive integers using the Fundamental Theorem of
Arithmetic in earlier classes, without realizing it! This method is also called the prime
factorization method. Let us recall this method through the following example.
Example-4. Find the HCF and LCM of 12 and 18 by the prime factorization method.
Solution : We have 12 = 2 ≥ 2 ≥ 3 = 22
≥ 31
18 = 2 ≥ 3 ≥ 3 = 21
≥ 32
Note that HCF (12, 18) = 21
≥ 31
= 6 = Product of thesmallest powerof each
common prime factor of the
numbers.
LCM (12, 18) = 22
≥ 32
= 36 = Product of the greatest powerof each
prime factor of the numbers.
Fromthe example above, you might have noticed that HCF (12, 18) ≥ LCM[12, 18]
= 12 ≥ 18. In fact, we can verify that for any two positive integers a and b,
HCF (a, b) ≥ LCM [a, b] = a ≥ b. We can use this result to find the LCM of two positive
integers, if we have already found the HCF of the two positive integers.
DO THIS
FindtheHCFandLCMof thefollowinggivenpairsofnumbersbyprimefactorisation
method.
(i) 120, 90 (ii) 50, 60 (iii) 37, 49
TRY THIS
Show that 3n
´ 4m
cannot end with the digit 0 or 5 for any natural numbers
‘n’and 'm'

RealNumbers 9
EXERCISE - 1.2
1. Express eachofthe following numbers as a product ofits prime factors.
(i) 140 (ii) 156 (iii) 3825 (iv) 5005 (v) 7429
2. Find the LCMand HCF ofthefollowing integers bytheprime factorizationmethod.
(i) 12, 15 and 21 (ii) 17, 23, and 29 (iii) 8, 9 and 25
(iv) 72 and 108 (v) 306 and 657
3. Check whether 6
n
canend with the digit 0 for anynaturalnumber n.
4. Explain why 7 ≥ 11 ≥ 13 + 13 and 7 ≥ 6 ≥ 5 ≥ 4 ≥ 3 ≥ 2 ≥ 1 + 5 are composite
numbers.
5. How will you show that (17 ≥ 11 ≥ 2) + (17 ≥ 11
1 ≥ 5) is a composite number?
Explain.
6. What is the last digit of 6
100
.
Now, let us use the Fundamental Theorem of Arithmetic to explore real numbers
further. First, we apply this theorem to find out when the decimal from of a rational number
is terminating and when it is non-terminating, repeating. Second, we use it to prove the
irrationality of many numbers such as 2 , 3 and 5 .
1.2.1 RATIONAL NUMBERS AND THEIR DECIMAL EXPANSIONS
Till now we have discussed some properities of integers. How can you find the
preceedingorthesucceedingintegersforagiveninteger?Youmighthaverecalledthatthedifference
between an integer and its preceeding or succeding integer is 1. And bythis propertyonlyyou
might havefoundrequired integers.
IncalssIX, youlearnedthat therationalnumberswouldbeineither aterminating decimal
formor a non-terminating, repeating decimalform. Inthis section, we are going to consider a
rational number, say
p
q
(q ¹ 0) and explore exactly when the number
p
q
is a terminating
decimal,andwhenit isanon-terminatingrepeating(orrecurring)decimal.Wedo sobyconsidering
certain examples
Let usconsiderthefollowing terminating decimals.
(i) 0.375 (ii) 1.04 (iii) 0.0875 (iv) 12.5

Class-XMathematics
10
Now let us express them in
p
q
form.
(i) 3
375 375
0.375
1000 10
< < (ii) 2
104 104
1.04
100 10
< <
(iii) 4
875 875
0.0875
10000 10
< < (iv) 1
125 125
12.5
10 10
< <
We see that all terminating decimals taken by us can be expressed in
p
q
formwhose
denominators are powers of10. Let us now factorize the numerator and denominator and then
expressthenin the simplest form:
Now (i)
3
3 3 3 3
375 3 5 3 3
0.375
10 2 5 2 8
≥
< < < <
≥
(ii)
3
2 2 2 2
104 2 13 26 26
1.04
10 2 5 5 25
≥
< < < <
≥
(iii)
3
4 4 4 4
875 5 7 7 7
0.0875
10 2 5 2 5 80
≥
< < < <
≥ ≥
(iv)
3
125 5 25
12.5
10 2 5 2
< < <
≥
Have you observed anypattern inthe denominators of the above numbers?It appears
that whenthe decimalisexpressed inits simplest rationalformthen pand q are coprime and the
denominator (i.e., q) hasonlypowers of2, or powers of 5, or both. Thisis because 2 and5 are
the onlyprime factors ofpowers of10.
Fromthe above examples, you have seenthat anyrationalnumber that terminates inits
decimal form can be expressed in a rational form whose denominator is a power of 2 or 5 or
both. So,whenwewritesucharationalnumber,in
p
q
form, theprime factorizationofq willbe in
2n
5m
, where n, m are some non-negative integers.
We canwriteour result formally:

RealNumbers 11
Theorem-1.3 : Let x be a rational number whose decimal form terminates. Then x can
be expressed in the form of
p
q
, where p and q are coprime, and the prime factorization
of q is of the form 2n
5m
, where n, m are non-negative integers.
DO THIS
Write the following terminating decimals in the form of
p
q
, q¹ 0 and p, q are
co-primes
(i) 15.265 (ii) 0.1255 (iii) 0.4 (iv) 23.34 (v) 1215.8
Write the denominators in 2n
5m
form.
You are probably wondering what happens the other way round. That is, if we have
a rational number in the form of
p
q
and the prime factorization of q is of the form 2n
5m
,
where n, m are non-negative integers, then does
p
q
have a terminating decimal expansion?
So, it seems to make sense to convert a rationalnumber ofthe form
p
q
,where q isofthe
form2n
5m
, to anequivalent rationalnumber ofthe form
a
b
, where b isa power of 10. Let us go
back to our examples above and work backwards.
(i)
3
3 3 3 3
3 3 3 5 375
0.375
8 2 2 5 10
≥
< < < <
≥
(ii)
3
2 2 2 2
26 26 13 2 104
1.04
25 5 2 5 10
≥
< < < <
≥
(iii)
3
4 4 4 4
7 7 7 5 875
0.0875
80 2 5 2 5 10
≥
< < < <
≥ ≥
(iv)
3
25 5 125
12.5
2 2 5 10
< < <
≥
So, these examples show us how we can convert a rational number of the form
p
q
,
where q is of the form 2n
5m
, to an equivalent rational number of the form
a
b
, where b is a
power of 10. Therefore, the decimal forms of such a rational number terminate. We find
that a rational number of the form
p
q
, where q is a power of 10, is a terminating decimal.
So, we conclude that the converse of theorem 1.3 is also true which can be formally
stated as :

Class-XMathematics
12
Theorem 1.4 : Let x =
p
q
be a rational number, such that the prime factorization of q is
of the form 2n
5m
, where n and m are non-negative integers. Then x has a decimal
expansion which terminates.
DO THIS
Write the denominator ofthefollowing rationalnumbers in 2n
5m
formwhere n and m
are non-negative integersand thenwritethemintheirdecimalform
(i)
3
4
(ii)
7
25
(iii)
51
64
(iv)
14
25
(v)
80
100
1.2.2 NON-TERMINATING, RECURRING DECIMALS IN RATIONAL
NUMBERS
Let us now consider rationalnumbers whose decimalexpansions are
non-terminatingand recurring.
Let uslook at the decimalconversion of
1
7
.
1
7
= 0.1428571428571 ..... whichis a non-terminating and recurring
decimal. Notice, the block ofdigits '142857' is repeating inthe quotient.
Notice that the denominator i.e., 7 can't be written in the form 2n
5m
.
DO THIS
Write thefollowing rationalnumbersinthe decimalformandfind out
the blockof repeating digits inthe quotient.
(i)
1
3
(ii)
2
7
(iii)
5
11
(iv)
10
13
Fromthe'Do this'exerciseandfromtheexampletakenabove, wecan
formallystate:
Theorem-1.5 : Let x =
p
q
be arational number, such that the primefactorization of q is
not of the form 2n
5m
, where n and m are non-negative integers. Then, x has a decimal
expansion which isnon-terminating repeating (recurring).
From the above discussion, we can conclude that the decimal form of every rational
numberiseither terminatingor non-terminating repeating.
0.1428571
7 1.0000000
7
30
28
20
14
60
56
40
35
50
49
10
7
30

RealNumbers 13
Example-5. Usingtheabovetheorems,withoutactualdivision,statewhetherdecimalformofthe
followingrationalnumbersareterminatingornon-terminating,repeatingdecimals.
(i)
16
125
(ii)
25
32
(iii)
100
81
(iv)
41
75
Solution : (i) 3
16 16 16
125 5 5 5 5
< <
≥ ≥
hasa terminating decimalform.
(ii) 5
25 25 25
32 2 2 2 2 2 2
< <
≥ ≥ ≥ ≥
hasaterminating decimalform.
(iii) 4
100 100 10
81 3 3 3 3 3
< <
≥ ≥ ≥
has a non-terminating, repeating decimalform.
(iv) 2
41 41 41
75 3 5 5 3 5
< <
≥ ≥ ≥
has anon-terminating, repeating decimalform.
Example-6. Writethedecimalformofthefollowingrationalnumberswithoutactualdivision.
(i)
35
50
(ii)
21
25
(iii)
7
8
Solution : (i) 1
35 7 5 7 7
0.7
50 2 5 5 2 5 10
≥
< < < <
≥ ≥ ≥
(ii)
2
2 2 2 2
21 21 21 2 21 4 84
0.84
25 5 5 5 5 2 5 2 10
≥ ≥
< < < < <
≥ ≥ ≥ ≥
(iii)
∋ ( ∋ ( ∋ (
3
3 3
3 3 3
7 7 7 7 5 7 25 875
0.875
8 2 2 2 2 2 5 2 5 10
≥ ≥
< < < < < <
≥ ≥ ≥ ≥
EXERCISE - 1.3
1. Write the following rational numbers in their decimal form and also state which
are terminating and which are non-terminating, repeating decimal.
(i)
3
8
(ii)
229
400
(iii)
1
4
5
(iv)
2
11
(v)
8
125
2. Without performing division, state whether the following rational numbers will
have a terminating decimal form or a non-terminating, repeating decimal form.
(i)
13
3125
(ii)
11
12
(iii)
64
455
(iv)
15
1600
(v)
29
343
(vi) 3 2
23
2 5
×
(vii) 2 7 5
129
2 5 7
× ×
(viii)
9
15
(ix)
36
100
(x)
77
210

Class-XMathematics
14
3. Write the followingrationals indecimalformusingTheorem1.4.
(i)
13
25
(ii)
15
16
(iii) 3 2
23
2 .5
(iv) 2 2
7218
3 .5
(v)
143
110
4. Express the followingdecimals inthe formof
p
q
, and writethe primefactorsofq.What
do you observe?
(i) 43.123 (ii) 0.1201201 (iii) 43.12 (iv) 0.63
1.3 IRRATIONAL NUMBERS
In classIX, you were introduced to irrational numbers and some of their properties.
You studied about their existence and how the rationals and the irrationals together made
up the real numbers. You even studied how to locate irrationals on the number line.
However, we did not prove that they were irrationals. In this section, we will prove that
2, 3, 5 and p in general is irrational, where p is a prime. One of the theorems, we
use in our proof, is the fundamental theorem of Arithmetic.
Recall, a real number is called irrational if it cannot be written in the form
p
q
,
where p and q are integers and q ¹ 0. Some examples of irrational numbers, with which
you are already familiar, are :
2, 3, 15, ,
ο 0.10110111011110…, etc.
Before we prove that 2 is irrational, we will look at a theorem, the proof of
which is based on the Fundamental Theorem of Arithimetic.
Theorem-1.6 : Let p be a prime number. If p divides a2
, (where a is a positive integer),
then p divides a.
Proof : Let the prime factorization of a be as follows :
a = p1
p2
… pn
, where p1
, p2
, …., pn
are primes, not necessarily distinct.
Therefore a2
= (p1
p2
… pn
) (p1
p2
… pn
) = p2
1
p2
2
… p2
n
.
Now, we are given that p divides a2
. Therefore, from the Fundamental Theorem of
Arithmetic, it follows that p isoneofthe primefactorsofa2
. However, usingtheuniquenesspart
of the Fundamental Theorem ofArithmetic, we realise that the only prime factors of a2
are
p1 , p2 ,… pn. So p is one of p1, p2, … pn.
Now, since a = p1p2 … pn, p divides a.

RealNumbers 15
DO THIS
Verify the theorem proved above for p= 2, p = 5 and for a2
= 1, 4, 9, 25, 36, 49, 64
and 81.
We are now readyto give a proofthat 2 is irrational. We will use a technique called
proofbycontradiction.
Example 7. Show that 2 is irrational.
Solution : Let us assume that 2 is rational.
If it is rational, then there must exist two integers r and s (s ¹ 0) such that 2 =
r
s
.
If r and s have a common factor other than 1, then, we divide r and s bytheir highest
common factor to get 2 =
a
b
, where a and b are co-prime. So, b 2 = a.
On squaring both sides and rearranging, we get 2b2
= a2
. Therefore, 2 divides a2
.
Now, byTheorem1.6, it follows that since 2 is dividing a2
, it also divides a.
So, we can write a = 2c for some integer c.
Substituting for a, we get 2b2
= 4c2
, that is, b2
= 2c2
.
This means that 2 divides b2
, and so 2 divides b (againusingTheorem1.6 withp=2).
Therefore, both a and b have 2 as a common factor.
But this contradicts the fact that a and bare co-prime.
This contradiction hasarisenbecause ofour assumption that 2 is rational. Thus our
assumptionis false. So, we conclude that 2 isirrational.
In general, it can be shown that d is irrational whenever d is a positive integer
which is not the square of another integer. As such, it follows that 6, 8, 15 24 etc.
are all irrational numbers.
In class IX, we mentioned that :
• the sum or difference of a rational and an irrational number is irrational and
• the product or quotient of a non-zero rational and irrational number is irrational.
We prove some particular cases here.

Class-XMathematics
16
Example-8. Show that 5 – 3 is irrational.
Solution : Let us assume that 5 – 3 is rational.
That is, we can find coprimes a and b (b ¹ 0) such that 5 – 3 =
a
b
.
Therefore, 5 –
a
b
= 3
we get 3 = 5
a
b
,
Since a and b are integers, 5
a
b
, is rational, and 3 is also rational.
But this contradicts the fact that 3 is irrational.
This contradiction has arisen because of our assumption that 5 – 3 is rational.
So, we conclude that 5 – 3 is irrational.
Example-9. Show that 3 2 is irrational.
Solution : Let us assume, the contrary, that 3 2 is rational.
i.e., we can find co-primes a and b (b ¹ 0) such that 3 2 =
a
b
.
we get 2 =
3
a
b
.
Since 3, a and b are integers,
3
a
b
is rational, and so 2 is rational.
But this contradicts the fact that 2 is irrational.
So, we conclude that 3 2 is irrational.
Example-10. Prove that 2 + 3 is irrational.
Solution : Let us suppose that 2 + 3 is rational.
Let 2 + 3 =
a
b
, where a, b are integers and b ¹ 0
Therefore, 2 =
a
b
– 3 .

RealNumbers 17
Squarring on bothsides, we get
2
2
2 3 2 3
a a
b b
< ∗ ,
Rearranging
2
2
2
3 3 2
a a
b b
< ∗ ,
<
2
2
1
a
b
∗
2 2
3
2
a b
ab
∗
<
Since a, b are integers,
2 2
2
a b
ab
∗
is rational, and so 3 is rational.
This contradicts the fact that 3 is irrational. Hence 2 3
∗ is irrational.
Note :
1. The sum of two irrational numbers need not be irrational.
For example, if a = 2 and b = 2
, , then both a and b are irrational, but a + b = 0
whichisrational.
2. The product oftwo irrationalnumbers neednot be irrational.
For example, a = 2 and b =3 2 , where both a and b are irrational, but
ab = 6 which is rational.
EXERCISE - 1.4
1. Prove that the following are irrational.
(i)
1
2
(ii) 3 + 5 (iii) 6 + 2 (iv) 5 (v) 3 + 2 5
2. Prove that p q
∗ is an irrational, where p, q are primes.
1.4 EXPONENTIALS REVISTED
We know the power 'an
' ofa number 'a' with natural exponent 'n' is the product of 'n'
factors each of which is equalto 'a' i.e
factors
-
= × × ××××××
14
4
244
3
n
n
a a a a a
20
, 21
, 22
, 23
.............. are powers of 2
30
, 31
, 32
, 33
............... are powers of 3

Class-XMathematics
18
We also know that when 81 is written as 34
, it is said to be in exponential form. The
number '4' is the 'exponent' or 'index' and 3 is the 'base'. we read it as " 81 is the 4th power of
base 3".
Recall the laws of exponents
If a, b are real numbers, where a ¹ 0, b ¹ 0 and m, n are integers, then
(i) am
. an
= am+n
; (ii) (ab)m
= am
.bm
(iii)
æ ö
=
ç ÷
è ø
m m
m
a a
b b
(iv) (am
)n
= amn
(v) a0
= 1 (vi) a–m
=
1
m
a
DO – THIS
1. Evaluate
(i) 21
(ii) (4.73)0
(iii) 03
(iv) (–1)4
(v) (0.25)–1
(vi)
2
5
4
æ ö
ç ÷
è ø
(vii)
2
1
1
4
æ ö
ç ÷
è ø
2. (a) Express10, 100, 1000, 10000, 100000 is exponential form
(b) Express insimplest exponentialform
(i) 16´ 64 (ii) 25´125 (iii) 128 ¸ 32
EXPONENTIAL AND LOGARITHIMS
Let usObservethe following
2x
= 4 = 22
gives x = 2
3y
= 81 = 34
gives y = 4
10z
= 100000 = 105
gives z = 5
Can we find the values ofxfor the following?
2x
= 5, 3x
= 7 , 10x
= 5
If so, what are the values of x?
For 2x
= 5, What should be the power to which 2 must be raised to get 5?
Therefore, we need to establisha new relation between x and 5.

RealNumbers 19
In suchsituation, a new relationlogarithm is introduced.
Consider y= 2x
, we need that value of x for which y becomes 5 from the facts that if
x = 1 then y = 21
= 2, if x = 2 then y = 22
= 4 , if x = 3 then y = 23
= 8, we observe that x
lies between 2 and 3.
We will now use the graph of y=2x
to locate such a 'x' for which 2x
= 5.
GRAPH OF EXPONENTIAL 2X
Let us draw the graph of y = 2x
For this we compute the value of 'y' bychoosing some values for 'x'.
x –3 –2 –1 0 1 2 3
y=2x
1
8
1
4
1
2
1 2 4 8
We plot the points and connect themwithsmooth curve.
Note that as xincreases, the
value of y = 2x
increases. As 'x'
decreases the value of
y = 2x
decreases veryclose to 0, but
never attains the value 0.
Let us think, if y = 2x
then
for which value ofx, y becomes 5?
We know that, in the graph
Y- axis represents the value of 2x
and X- axis represents the value of
x. Locate the value of 5 onY- axis,
and represent it as a corresponding
point "P" on Y- axis. Draw a line
parallelto X- axis throughP, which
meets the graphat the point Q.
Now draw QR perpendicular to X - axis. Can we find the length ofOR approximately
fromthe graph?or where does it lie?Thus, we knowthat the x coordinateofthe point Ris the
required value of x, for which 2x
=5.
This value of x is called the logarithmof 5 to the base 2, writtenas log2
5.
The Curve comes
closer to the X-axis,
but neither touch nor
cross the X-axis.
-5
X'
P Q
5
3
1
7
9
O 1 3 5
-1
-3 R X
Y'
Y'

Class-XMathematics
20
THINK AND DISCUSS
Let us observe the scale factor in the graph of y = 2x
On X- axis (Refer Ratio - proportion)
If 10 places = 1 unit
20 places= 2 units
40 places = 4 units, then
CanyouimaginethecorrespondingvalueonX-axis,withreferencetothe5onY-axis?
We rewrite the above table as follows
x –2 –1 0 1 2 3 y
1
4
1
2
1 2 4 8
y = 2x
1
4
1
2
1 2 4 8 x = log2
y –2 –1 0 1 2 3
Observe the graph y = 2x
inthe light ofourdefinitionoflogarithm
If y =
1
4
; x = – 2 i.e. 2–2
=
1
4
and –2 = log2
1
4
y =
1
2
; x = –1 i.e. 2–1
=
1
2
and –1 = log2
1
2
y = 2 ; x = 1 i.e. 21
= 2 and 1 = log2
2
y = 4 ; x = 2 i.e. 22
= 4 and 2 = log2
4
y = 8 ; x = 3 i.e. 23
= 8 and 3 = log2
8
i.e. y – cordinate of anypoint on the curve is the xth power of2, and x – coordinate of
anypoint onthe curveis the logarithmof y – coordinate to the base 2.
Let us considerone more example :
If 10 y
= 25 then it can be represented as y = log10
25 or y = log 25,
Logarithms ofa number to thebase 10 are also called commonlogarithms. Inthis case,
we generallyomit the base i.e. log10
25 is also written as log 25.
In general, a and N are positive real numbers such that 1
a ¹ we define
loga
N = x Û ax
= N.

RealNumbers 21
DO – THIS
(1)Writethefollowing inlogarithmic form.
(i) 7 = 2x
(ii) 10 = 5b
(iii)
1
3
81
c
= (iv) 100 = 10z
(iv)
1
4
257
a
=
(2)Write thefollowing inexponentialform.
(i) log10
100 = 2 (ii) log5
25 = 2 (iii) log2
2 = 1
TRY THIS
Solve thefollowing
(i) log2
32 = x (ii) log5
625 = y (iii) log10
10000 = z
(iv) logx
16 = 2 x2
= 16 Þ x = 4
± , Is it correct or not?
Can we say "exponential formand logarithm" formare inverses ofone another?
Also, observe that everypositive realnumber has a unique logarithmic value, because
anyhorizontallinetouchesthe graph at onlyone point.
THINK AND DISCUSS
(1) Does log2
0 exist? Give reasons.
(2) Prove (i) logb
b = 1 (ii) logb
1 = 0 (iii) logx
bx
= x
PROPERTIES OF LOGARITHMS
Logarithms aremore important in manyapplications, and also inadvanced mathematics.
Wenow estiablishsome basic propertiesusefulin manipulating expressionsinvolving logarithms.
(i) The Product Rule
The propertiesofexponents correspond to properties oflogarithms. Forexample when
we multiplywith the same base, we add exponents
i.e. ax
. ay
= ax+y
This propertyofexponents coupled with an awareness that a logarithmis an exponent
suggest the Product Rule.
Theorem: (Product Rule) Let a, x and y be positive real numbers with a ¹ 1.
Then loga
xy = loga
x + loga
y
i.e., The logarithm of a product is the sum of the logarithms
Proof:
Let loga
x = m and loga
y = n then we have am
= x and an
= y

Class-XMathematics
22
Now
xy = am
an
= am+n
loga
xy = m + n = loga
x + loga
y
TRY THIS
We know that log10
100000 = 5
Show that you get the same answer bywriting 100000 = 1000 ´ 100 and then using
the product rule.Verifythe answer.
DO THIS
Expressthe logarithms ofthefollowing as the sumofthe logarithm
(i) 35 ´ 46 (ii) 235 ´ 437 (iii) 2437 ´3568
(ii) The Quotient Rule
When we divide with the same base, we subtract exponents
i.e.
x
x y
y
a
a
a
-
=
This propertysuggests the quotientrule.
Theorem: (Quotient Rule) Let a, x and y be positive real numbers where a ¹ 1.
Then loga
x
y
= loga
x– loga
y
i.e. the logarithm of a quotient is the difference of the logarithms of the two
numbers taken in the same order.
TRY THIS
We know log2
32 = 5. Showthat we get the same answer bywriting 32 as
64
2
andthen
using theproduct rule. Verifyyour answer .
DO THIS
Expressthe logarithmsofthe following asthedifference oflogarithms
(i)
23
34
(ii)
373
275
(iii) 4325 ¸ 3734 (iv) 5055 ¸ 3303

RealNumbers 23
THINK AND DISCUSS
Prove the quotient ruleusing
m
m n
n
a
a
a
-
= .
(iii) The Power Rule
When anexponentialexpressionis raised to a power, wemultiplythe exponents
i.e. (am
)n
= am.n
This propertysuggests the power rule.
Theorem: (Power Rule) Let a and x be positive real numebrs with a ¹ 0 and n
be any real number
then, loga
xn
= n loga
x
i.e. the logarithm of a number with an exponent is the product of the exponent
and the logerithmof that number.
TRY THIS
We have log2
32 = 5.Show that we get the same result bywriting32 = 25
and then using
power rules.Verify the answer.
Can we find the value of x such that 2x
= 35
? In such cases we find the value of
35
= 243. Then we can evaluate the value of x, for which the value of2x
equals to 243.
Applyingthelogarithmandusing theformula log log
a
n
a
x n x
= ,Easilywecanfind the
values of 325
, 333
etc.
2x
= 35
writing in logarthmic form
5
2
log 3 x
=
2
5 log 3 x
= ( )
log log
n
a a
x n x
=
Q
We observe that the value of x is the product of 5 and the value of 2
log 3.
DO THIS
Using loga
xn
= n loga
x, expand the following
(i) log2
725
(ii) logs5
850
(iii) log 523
(iv) log1024
Note: log x = log10
x

Class-XMathematics
24
THINK AND DISCUSS
We can write log
x
y
= log (x.y–1
). Can you prove that log
x
y
= log x – logy using
product and power rules.
TRY THIS
(i) Find the value oflog2
32 (ii) Find the value of logc c
(iii) Find the value of log10
0.001 (iv) Find the value of 2
3
8
log
27
THINK - DISCUSS
We know that,if 7 = 2x
then x=log2
7. Then, what is the value of 2
log 7
2 ?Justifyyour
answer. Generalise the above bytaking some more examples for log N
a
a
Example-11. Expand log
343
125
Solution : As you know, loga
x
y
= loga
x - loga
y
So, log
343
125
= log343 – log125
= log73
– log53
= 3log7 – 3log5 ( Since, loga
xn
= m loga
x )
So log
343
125
= 3(log7 – log5).
Example-12. Write 2log3 + 3log5 – 5log2 as a single logarithm.
Solution : 2log3 + 3log5 – 5log2
= log32
+ log53
– log25
( Since in n loga
x=loga
xn
)
= log9 + log125 – log32
= log (9 × 125) – log32 ( Since loga
x + loga
y = loga
xy )
= log1125 – log32
= log
1125
32
(Since loga
x – loga
y = loga
x
y
)

RealNumbers 25
Example-13. Solve 3x
= 5x-2
.
Solution : x log10
3 = (x - 2) log10
5
x log10
3 = xlog10
5 - 2log10
5
xlog10
5 - 2log10
5 = x log10
3
xlog10
5 - x log10
3 = 2log10
5
x(log10
5 - log10
3) = 2log10
5
x = 10
10 10
2log 5
log 5 log 3
-
Example-14. Find x if 2log5 +
1
2
log9 - log3 = log x
Solution : logx = 2log5 +
1
2
log9 - log3
= log52
+ log
1
2
9 - log3
= log25 + log 9 - log3
= log25 + log3 - log3
log x = log25
x = 25
EXERCISE - 1.5
1. Determine the value of the following.
(i) log25
5 (ii) log81
3 (iii) log2
1
16
æ ö
÷
ç ÷
ç ÷
ç
è ø
(iv) log7
1 (v) logx x (vi) log2
512
(vii) log10
0.01 (viii) 2
3
8
log
27
æ ö
÷
ç ÷
ç ÷
ç
è ø
(ix) 2
2 log 3
2 +
2. Write the following expressionsaslog N and find their values.
(i) log 2 + log 5 (ii) log2
16 - log2
2 (iii) 3 log64
4
(iv) 2 log 3 - 3 log 2 (v) log 10 + 2 log 3 - log 2

Class-XMathematics
26
3. Evaluate each of the following in terms of x and y, if it is given that x = log2
3 and
y = log2
5
(i) log2
15 (ii) log2
7.5 (iii) log2
60 (iv) log2
6750
4. Expand thefollowing.
(i) log1000 (ii) log
128
625
æ ö
÷
ç ÷
ç ÷
ç
è ø
(iii) logx2
y3
z4
(iv) log
2 3
4
p q
r
æ ö
ç ÷
è ø
(v) log
3
2
x
y
5. If x2
+ y2
= 25xy, then prove that 2 log(x + y) = 3log3 + logx + logy.
6. If
1
log (log log )
3 2
x y
x y
æ ö
∗ ÷
ç < ∗
÷
ç ÷
ç
è ø
, thenfind the value of
x y
y x
∗ .
7. If (2.3)x
= (0.23)y
= 1000, then find the value of
1 1
x y
, .
8. If 2x+1
= 31-x
then find the value of x.
9. Is (i) log 2 rationalor irrational?Justifyyour answer.
(ii) log 100 rationalor irrational? Justifyyour answer.
OPTIONAL EXERCISE
[For extensive learning]
1. Can the number 6n
, nbeing a naturalnumber, end withthedigit 5?Give reason.
2. Is 7 ≥ 5 ≥ 3 ≥ 2 + 3 a composite number? Justify your answer.
.
3. Prove that ∋ (
2 3 5
∗ is an irrational number. Also check whether
∋ (∋ (
2 3 5 2 3 5
∗ , is rationalor irrational.
4. If x2
+ y2
= 6xy, prove that 2 log (x + y) = logx + logy + 3 log 2
5. Find the number of digits in 42013
, iflog10
2 = 0.3010.
Note : Ask your teacher about integralpart and decimalpart of the logarithm of a number.

RealNumbers 27
WHAT WE HAVE DISCUSSED
1. Division Algorithm: Given positive integers a and b, there exist whole numbers
q and r satisfying a = bq + r, 0 < r < b.
2. The FundamentalTheorem ofArithmetic states that every composite number can be
expressed (factorized) as a product of itsprimes, and this factorizationis unique, apart
fromthe order inwhichtheprime factors occur.
3. If p is a prime and p divides a2
, where a is a positive integer, then p divides a.
4. Let x be arationalnumber whose decimalexpansion terminates. Then wecanexpress x
in the formof
p
q
, where p and q are coprime, and the prime factorization ofq is ofthe
form2n
5m
, where n and m are non-negative integers.
5. Let x =
p
q
be a rational number, such that the prime factorization of q is of the form
2n
5m
, where n and m are non-negative integers. Then x has a decimalexpansion which
terminates.
6. Let x =
p
q
be arationalnumber, such that the prime factorizationofq is not ofthe form
2n
5m
, where n and m are non-negative integers. Then x has a decimalexpansion which
isnon-terminatingand repeating(recurring).
7. We define loga
x = n, if an
= x, where a and x are positive numbers and a ¹ 1.
8. Lawsof logarithms :
If a, x and y are positive real numbers and a ¹ 1, then
(i) loga
xy = loga
x + loga
y (ii) loga
x
y
= loga
x , loga
y
(iii) loga
xm
= m loga
x (iv) log N
a
a = N (v) loga
1 = 0
(vi) loga
a = 1
9. Logarithmsare used for calculations inengineering, science, business and economics.
Suggested Projects
Euclid Algorithm
l Find the H.C.FbyEuclid Algorithmbyusing colour ribbon band or grid paper.

2.1 INTRODUCTION
How would you describe a person, when you are asked to do?
Let us see some examples.
Ramanujan was a mathematican, interested in numbertheory.
Dasarathi was a telugu poet, and also a freedom fighter.
Albert Eienstein German by birth was a physicist and music was his hobby.
Maryam Mirzakhan is the only women mathematican to win Fields medal.
We classify the individuals with specific character and interest, then as a member of
larger more recogizable group. People classifyand categorythe world around inorder to make
sense oftheir environment and their relationship to other.
Books in the library are arranged according to the subject, so that we can find the
required quickly.
In chemistrythe elements are categorized in groups and classes to studytheir general
properties
Your mathematics syllabus for tenth class has been divided into 14-Chapters under
different headings.
DENTAL FORMULA
Observe set of human teeth is of
classified into 4-types according to
their functions.
(i) Incisors (ii) Canines
(iii) Premolars and (iv) Molars
For each quadrant, for the sequence
of incisors, canines, premolars and
molarswe writethe dentalformula2,
1, 2, 3 specifying their number.
Sets
2

29
Mathematicsisno different fromothers,itneeds to collect objectsinto meaningfulgroups
Among such, groupsofnumbers we commonlyuseinmathematics are
¥ = Collection ofnatural numbers 1, 2, 3....
W = Collection of whole numbers 0, 1, 2, 3.....
I or ¢ = Collection of Integers 0, ± 1, ± 2, ± 3, ......
¤ = Collection of rational numbers i.e the numbers that can be written in
p
q
form
where p, q are integers and q ¹ 0
¡ = Collection of real numbers i.e. the numbers which have decimal expansion.
DOTHIS
Identifyand writethe “commonproperty” ofthe following collections.
1) 2,4,6,8,… 2) 2,3,5,7,11,… 3) 1,4,9,16,…
4) January, February, March,April,…
5)Thumb,indexfinger, middlefinger,ringfinger, pinky
THINK AND DISCUSS
Observe the following collections and prepare as manyas generalized statements you
can describing their more properties.
1) 2,4,6,8,… 2) 1,4,9,16,…
However, we sometimescome across certain groups, inwhich the objects do not share
anycommon property.
For example,
2, Ramesh, January
Theremaybe somereasonforforming thisgroup, butwe don’t seeanycommonproperty
shared bythese objects. Nevertheless, we are sure that thereare onlythree objects inthis group
and we also know what these objects are. In other words, these objects are ‘well defined’.
29

Class-XMathematics
30
2.2 SET
A set is a well defined collection of distinct objects. The objects in a set are
called elements. Sets are written by enclosing all its elements between the brackets { }.
For example, whenwe want to writea set offirst fiveprime numbers, it canbe writtenas
{2,3,5,7,11} and set ofincisors ={centralincisor, lateralincisor}
DOTHIS
Writethefollowing sets.
1) Set ofthefirst five positive integers.
2) Set ofmultiplesof 5 which are more than 100 and lessthan 125
3) Set offirst five cubic numbers.
4) Set ofdigitsinthe Ramanujannumber
2.2.1 ROSTER FORM AND SET BUILDER FORM
It isdifficult to express a set in a long sentence. Therefore, sets are generallydenoted by
capitalletters ofEnglishalphabetA, B, C.....
For example, M is the set of molars among our teeth.
We can write this set asM={first molar, second molar, third molar}.
Let us look at another example. Q is the set of quadrilaterals with at least two equal
sides. Then, we can write this set as
Q ={square, rectangle, rhombus, parallelogram, kite, isosceles trapezium, dart}
Here, we are writing a set bylisting the elements in it. In suchcase, the set is said to be
written inthe “rosterform”.
In the above two examples, let us discuss belongingness ofthe elements and its repre-
sentation. Suppose, ifwe want to say“second molar is in the set ofmolars”, thenwe canrepre-
sent this as“second molar Î M”.And we read this as "second molar belongs to set M"
Canwe say“rhombus Î Q”inthe above example of set of quadrlaterals? Howdo you
read this?
Does “square” belong to the set M inthe above examples?
Then, how do we denote this?When we say“square is not inthe set M”, we denote as
“square Ï M”.And we read this as "square does not belong to the set M".
Recall from the classes you have studied earlier that we denote natural numbers by
,
¥ set ofintegers by ,
¢ set ofrationalnumbers by ,
¤ and set ofreal numbers by ¡ .

31
Sets
DOTHIS
Somenumbersaregivenbelow. Decide thenumbersto whichnumbersetstheybelong
to and does not belong to and express with correct symbols.
i) 1 ii) 0 iii) -4 iv)
5
6
v) 1.3
vi) 2 vii) log 2 viii) 0.03 ix) p x) 4
-
THINK AND DISCUS
Canyou writethe set ofrationalnumbersinroster form?
Youmighthaveconcluded byyourearlier discussionthat itisnot possibleto write theset
of rational numbers byshowing list ofelements in it.You might have also concluded that allthe
rationalnumbers are writteninthe formof
p
q
(q¹ 0 and p,qare integers).
When wewrite a set bydefining its elements witha “common property”, we cansaythat
the set is in the “set builder form”. Set builder formshould follow somesyntax. Let usknowit by
observing anexample.
SupposeAis a set ofmultiples of 3 lessthan 20. Then,A={3,6,9,12,15,18} and this is
roster formofthe setA.Whenwe write its set builder form, it is
A={x : x is amultiple of 3, x < 20} and we
read this as “Ais the set ofelements xsuch
that x isa multiple of3 and x islessthan20.
Similarly,wecanexpresstherationalnumberssetas ¤={x:x=
p
q
,p,qareintegersandq¹ 0}
Intheexample,
{2, Ramesh, January}
There are three objects forming a set. But theydo not share anycommonproperty. So
we can’t express it in the set builder form.
Note : (i) In roster form, the order in which the elements are listed is immaterial. Thus,
the set ofdigits in the Ramanujamnumber is {7, 2, 1, 9}
(ii) Whilewritingtheelementsofaset inroster form,anelement isnot repeated.For
example, theset ofletters formingthe word “SCHOOL” is{s, c, h, o,l} and not
{s, c, h, o, o, l}. Therefore a set contains distinct elements.
A={ x : x is a multiple of 3 and x<20}
the set all x such that x is a multiple of 3 and x<20

Class-XMathematics
32
Let us observe "roster form" and"set builder form" ofsome sets.
Roster form Set builder form
V = {a, e, i, o, u} V = {x : x is a vowel in the english alphabet}
A = {-2,-1,0,1,2} A = {x : -2£x£2, xÎ ¢ }
B = {1,
1 1 1 1
, , ,
2 3 4 5
} B = {x : x =
1
,
n
nÎ ¥ , n£5}
C = {2,5,10,17} C = {x : x = 2
1, , 4
n n n
+ Î £
¥ }
DOTHIS
1. List theelements ofthefollowing sets.
(i) G = {all the factors of 20}
(ii) F = {the multiples of 4 between 17 and 61 which are divisible by 7}
(iii) S = {x : x is a letter in the word 'MADAM'}
(iv) P = {x : x is a whole number between 3.5 and 6.7}
2. Write the following sets in the roster form.
(i) B is the set of all months in a year having 30 days.
(ii) P is the set of all prime numbers smaller than 10.
(iii) X is the set of the colours of the rainbow
3. A is the set of factors of 12. Which one of the following is not a member of A.
(A) 1 (B) 4 (C) 5 (D) 12
TRY THIS
1. Write some sets of your choice, involving algebraic and geometrical ideas.
2. Match roster forms with the set builder form.
(i) {p, r, i, n, c, a, l} (a) {x : x is a positive integer and is a divisor of 18}
(ii) {0} (b) {x : x is an integer and x2
– 9 = 0}
(iii) {1, 2, 3, 6, 9, 18} (c) {x : x is an integer and x + 1 = 1}
(iv) {3, -3} (d) {x : x is a letter of the word PRINCIPAL}

33
Sets
EXERCISE - 2.1
1. Whichofthe following are sets?Justifyyour answer.
(i) The collection ofallthe months ofa year begining withthe letter “J”.
(ii) The collection of tenmost talentedwriters ofIndia.
(iii) A teamofeleven best cricket batsmenofthe world.
(iv) The collectionofallboysinyour class.
(v) The collectionofalleven integers.
2. IfA={0, 2, 4, 6}, B = {3, 5, 7} and C = {p, q, r}, then fill the appropriate symbol,
Î or Ï in the blanks.
(i) 0 ….. A (ii) 3 ….. C (iii) 4 ….. B
(iv) 8 ….. A (v) p ….. C (vi) 7 ….. B
3. Expressthefollowingstatementsusingsymbols.
(i) The element ‘x’does not belong to ‘A’.
(ii) ‘d’ isan element ofthe set ‘B’.
(iii) ‘1’ belongs to the set ofNaturalnumbers.
(iv) ‘8’does not belong to the set ofprime numbers P.
4. State whether the following statements are true or false. Justifyyour answer
(i) 5 Ï set ofprime numbers
(ii) S = {5, 6, 7} implies 8 Î S.
(iii) -5 Ï W where‘W’ is the set ofwhole numbers
(iv)
8
11
Î ¢ where ‘¢ ’ is theset ofintegers.
5. Write the followingsetsin roster form.
(i) B = {x : x is a naturalnumber smaller than 6}
(ii) C = {x : x is atwo-digit naturalnumber such that the sumofits digits is 8}.
(iii) D = {x : x is a prime number which is a divisor of60}.
(iv) E = {x : x is analphabet in BETTER}.
6. Write thefollowing sets intheset-builder form.
(i) {3, 6, 9, 12} (ii) {2, 4, 8, 16, 32}
(iii) {5, 25, 125, 625} (iv) {1, 4, 9, 16, 25, ….. 100}
7. Write the followingsetsin roster form.
(i) A = {x : x is a naturalnumber greater than 50 but smaller than 100}
(ii) B = {x : x is an integer, x2
= 4}
(iii) D = {x : x is a letter in the word “LOYAL”}

Class-XMathematics
34
8. Match the roster formwithset builder form.
(i) {1, 2, 3, 6} (a) {x : x is prime number and a divisor of6}
(ii) {2, 3} (b) {x : x is an odd natural number smaller than 10}
(iii) {m, a, t, h, e, i, c, s} (c) {x : x is a natural number and divisor of6}
(iv) {1, 3, 5, 7, 9} (d) {x : x is a letter ofthe word MATHEMATICS}
2.3 EMPTY SET
Let usconsider the following examplesofsets:
(i) A = {x : x is a natural number smaller than 1}
(ii) D = {x : x is a odd number divisible by 2}
How many elements are there in setsAand D? We find that there is no natural number
whichissmaller than1. So setAcontains noelements or wesaythatAis anemptyset.Similarly,
there are no odd numbers that are divisible by 2. So, D is also an empty set.
Aset whichdoes not containanyelement is calledanempty set,or a Nullset, or a void
set. Emptyset is denoted bythe symbol f or { }.
Here aresome more examples ofempty sets.
(i) A = {x : 1 < x < 2, x is a natural number}
(ii) B = {x : x2
– 2 = 0 and x is a rational number}
Note : f and {0} are two different sets. {0} is a set containing an element 0 while f has no
elements (nullset).
2.4 UNIVERSAL SET AND SUBSET
Consider the teeth set that we have discussed in the
begining of the chapter. You have classified the whole teeth
set into four sets namely incisors, canines, premolars and
molars.
But, are teeth in the set of molars also members of whole
teeth set? or not?
Here, whole teeth set is "universalset" of abovesaid four teeth sets.
incisors canines
pre-molars molars
teeth set

35
Sets
Consider the teeth set as universal set and canines, incisors
are two sets then we can represent this as shown in the adjacent
diagram also.
Observe the diagram. What does the remaining empty
part of the diagram represents?
Let us see some more examples of universal sets:
(i) If we want to study the various groups of people of our state (may be according to
income or work or caste), universal set is the set of all people in Telangana.
(ii) If we want to study the various groups of people in our country, universal set is the
set of all people in India.
The universal set is generallydenoted by 'm' and sometimes by U. The Universal set is
usuallyrepresented byrectangles to showinthe formofa figure.
Let usconsidertheset ofnaturalnumbers,
¥ ={1,2,3,4...}.Then set ofeven numbers is
formed by the elements of ¥ . Then ¥ is
unversal set of even numbers. Is ¥ also
universalset for the set ofodd numbers?
Consider aset fromA={1,2,3}.Howmanysetscanyou formbytakingasmanyelements
as you wish from set A?
Now, {1},{2},{3},{1,2},{2,3},{1,3} and {1,2,3} are the sets you can form. Can you
form anyother sets?These sets are called subsetsof A. Ifwewant to say{1,2} issubset of A,
then we denote it as {1,2}Ì A. When we consider the subsets ofA, we should say {1,2,3} is
also as a subset ofA.
If allelements ofsetAarepresent in B, then Ais said to besubset of B denoted byAÍ B.
Then we can write as A Í B Û "a Î A Þ a Î B", whereA and B are two sets.
Let us consider the set ofreal numbers ¡ ; It has manysubsets.
For example, the set of naturalnumbers ¥ = {1, 2, 3, 4, 5, ……},
the set of whole numbers W = {0, 1, 2, 3, ......},
the set of integers ¢ = {…., -3, -2, -1, 0, 1, 2, 3, …..}
The set ofirrationalnumbers ¤'is composedofallreal numbers whichare not rational.
Consider a null set f and a non emptysetA. Is f a subset ofA? Ifnot f should have
anelement whichisnot element ofA. for beinganemptyset f hasno suchelement, thus f ÌA.
1. When we say " if x<3, then x<4",
we denote as "x<3 Þ x<4".
2. When we say "x-2 =5 if and only if x=7",
we write this as "x-2 = 5 Û x=7"
¥
¡
canines incisors
m

Class-XMathematics
36
Null set is a subset of every set.
Is AÍA?Allelements of LHS setAare also elements ofRHS setA.Thus … .
Every set is a subset of itself.
Thus, ¤'= {x : x Î ¡ and x Ï ¤ } i.e., allrealnumbers that are not rational. e.g. 2 ,
5 and p.
Similarly, the set ofnaturalnumbers, Nisa subset ofthe set of
wholenumbersWandwecanwrite ¥ Ì W.AlsoWisasubet of ¡ .
That is ¥ Ì W and W Ì ¡
Þ ¥ Ì W Ì ¡
Some of the obvious relations among these subsets are
¥ Ì ¢ Ì ¤ Ì ¡ and ¤' Ì ¡ , and ¥ Ë ¤'.
Consider the set of vowels,V = {a,e, i, o, u}.Also consider the setA,of alllettersin the
English alphabet. A= {a, b, c, d, ….., z}. We can see that every element of set V is also an
element A. But there are elements ofAwhich are not a part of V. In this case, Vis called the
proper subset ofA.
In other words V Ì A since, whenever a Î V, then a Î A It can also be denoted by
V A
Í and is read as V is the subset ofA.
DO THIS
1. A = {1, 2, 3, 4}, B = {2, 4}, C = {1, 2, 3, 4, 7}, F = { }.
Fill in the blanks with Ì or Ë .
(i) A ….. B (ii) C ….. A (iii) B ….. A
(iv) A ….. C (v) B ….. C (vi) f ….. B
2. State whichofthe following statement are true.
(i) { } = f (ii) f = 0 (iii) 0 = { 0 }
¥
¡
W
¥
W

37
Sets
TRY THIS
1. A={set ofquadrilaterals}, B={square,rectangle,trapezium,rhombus}. Statewhether
AÌ Bor B Ì A. Justifyyour answer.
2. IfA= {a, b, c, d}. How manysubsets does the set Ahave?
(A) 5 (B) 6 (C) 16 (D) 65
3. P isthe set offactorsof5, Qisthe set offactorsof25 and Ris the set offactorsof125.
Whichone ofthe following isfalse?
(A) P Ì Q (B) Q Ì R (C) R Ì P (D) P Ì R
4. Ais theset ofprime numberslessthan 10, B is the set ofodd numberslessthan10and
C is the set of even numbers lessthan 10. Whichof the following statementsare true?
(i)AÌ B (ii) B ÌA (iii)AÌ C
(iv) C ÌA (v) B Ì C (vi) f ÌA
A
2.5 VENN DIAGRAMS
We have already seen different ways of representing sets using diagrams. Let us learn
about Venn-Euler diagramor simplyVenn-diagram. It is one of the ways of representing the
relationships between sets. These diagrams consist of rectangles and closed curves usually
circles.
As mentioned earlier in the chapter, the universalset is
usuallyrepresented bya rectangle.
(i) Consider that m = {1, 2, 3, …., 10} is the universalset
for which, A = {2, 4, 6, 8, 10} is a subset. Then the
Venn-diagramis as:
(ii) m = {1, 2, 3, …., 10} is the universal set of which,A
= {2, 4, 6, 8, 10} and B = {4, 6} are subsets and
also B ÌA. Then, the Venn-diagramis :
(iii) Let A= {a, b, c, d} and B = {c, d, e, f}.
Then we illustrate these sets with aVenn diagramas
1 A 3
2
4
6
10
5
9
7
8
m
1
9
3
5
7
A
B
2
8
10
4
6
m
a
b
e
f
B
A
c
m

Class-XMathematics
38
2.6 BASIC OPERATIONS ON SETS
Weknowthat arithmeticshasoperationsofaddition,subtraction,multiplication anddivision
ofnumbers.Similarlyinsets, wedefinetheoperationofunion,intersectionanddifferenceofsets.
2.6.1 UNION OF SETS
Let us consider m , the set ofall studentsin your school.
Suppose Ais the set of students inyour class who were absent on Tuesday and B is the
set ofstudents who were absent on Wednesday. Then,
Let A= {Roja, Ramu, Ravi} and
Let B = {Ramu, Preethi, Haneef}
Now, we want to find K, the set of students who were absent on either Tuesday or
Wednesday. Then, does Roja Î K? Ramu Î K? Ravi Î K? Haneef Î K? Preethi Î K?
Akhila Î K?
Roja, Ramu, Ravi, Haneefand Preethiallbelong to
K butAkhila does not.
Hence, K= {Roja, Ramu, Ravi, Haneef, Preethi}
Here, the set K is the called the union ofsetsAand
B. The unionofAand B is the set which consists of all the
elements ofAand B. The symbol‘È’is used to denote theunion. Symbolically, we writeAÈ B
and usually read as ‘Aunion B’orAcup B.
A È B = {x : xÎA or xÎB}
Example-1. Let A= {2, 5, 6, 8} and B = {5, 7, 9, 1}. Find AÈ B.
Solution : We have AÈ B = {2, 5, 6, 8} È {5, 7, 9, 1}
= {2, 5, 6, 8, 5, 7, 9, 1}
= {1, 2, 5, 6, 7, 8, 9}.
Note that the commonelement 5 was taken onlyonce while writingAÈ B.
Example-2. Let A= {a, e, i, o, u} and B = {a, i, u}. Show that AÈ B = A.
Solution : We have AÈ B = {a, e, i, o, u} È {a, i, u}
= {a, e, i, o, u, a, i, u}
= {a, e, i, o, u} = A.
A B m

39
Sets
This example illustrates that union ofsetsAand its subset B is the set Aitself.
i.e, if B ÌA, then AÈ B =A.
Example-3. If A= {1, 2, 3, 4} and B = {2, 4, 6, 8}. Find AÈ B.
Solution : A= {1, 2, 3, 4} and B = {2, 4, 6, 8}
then AÈ B = {1, 2, 3, 4}È {2, 4, 6, 8}
= {1, 2, 3, 4, 2, 4, 6, 8}
= {1, 2, 3, 4, 6, 8}
2.6.2 INTERSECTION OF SETS
Let us again consider the example ofstudents who were absent.Now let usfind the set L
that represents the students who were absent on both Tuesday and Wednesday. We find that
L = {Ramu}.
Here, the set Lis called the intersection ofsetsAand B.
Ingeneral, theintersectionofsetsAandB is theset ofall
elementswhicharecommon inboth AandB.i.e.,thoseelements
whichbelongtoAand also belong to B. Wedenote intersection
symbolicallybyasAÇ B (read as “Aintersection B”).
i.e., A Ç B = {x : x Î Aand x Î B}
The intersection ofAand B can be illustrated using the Venn-diagram as shown in the
shaded portion ofthe figure, given below, for Example 5.
Example-4. Find AÇ B whenA= {5, 6, 7, 8} and B = {7, 8, 9, 10}.
Solution : The common elements in both Aand B are 7 and 8.
A Ç B = {5, 6, 7, 8} Ç {7, 8, 9, 10} = {7, 8}
(commonelements)
Example-5. IfA= {1, 2, 3} and B = {3, 4, 5}, thenillustrate
AÇ B inVenn-diagrams.
Solution : The intersection ofAand B can be illustrated in the
Venn-diagramas showninthe adjacent figure.
DISJOINT SETS
Suppose A = {1, 3, 5, 7} and B = {2, 4, 6, 8}. We see
that there are no common elements in Aand B. Such sets are
known as disjoint sets. The disjoint sets can be represented by
meansoftheVenn-diagramas follows:
A B
A Ç B = {3}
3
1
2
4
5
m
m
A B
1
3
2
4
6
8
A È B = {1, 2, 3, 4, 6, 8}
A
B
m
1 3
5 7
2 4
6 8
A Ç B = f
m
A Ç B
A B

Class-XMathematics
40
DO THIS
1. Let A= {1, 3, 7, 8} and B = {2, 4, 7, 9}. Find AÇ B.
2. IfA= {6, 9, 11}; B = { }, find AÈ f .
3. A = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}; B = {2, 3, 5, 7}. Find A Ç B.
4. IfA= {4, 5, 6}; B = {7, 8} then show that AÈ B = B È A.
TRY THIS
1. List out some setsAand B and choose their elements such that Aand B are disjoint
2. If A= {2, 3, 5}, findAÈ f and f È Aand compare.
3. If A= {1, 2, 3, 4} and B ={1, 2, 3, 4, 5, 6, 7, 8}, thenfindAÈ B and AÇ B. What
do you noticeabout the result?
4. Let A= {1, 2, 3, 4, 5, 6} and B = {2, 4, 6, 8, 10}. Find the intersection ofAand B.
THINK AND DISCUSS
The intersection ofanytwo disjoint setsis a nullset. Justifyyour answer.
2.6.3 DIFFERENCE OF SETS
SupposeAis the set ofnatural numbers less than10 and B isthe set of evennumbers less
than 10. Ifwe consider theset ofodd numbers less than 10, then the elements inthe set belong
to set A, but not to the set B. This set is represented byA- B and read it asAdifference B
A - B = {1, 3, 5, 7, 9}.
Now we define the difference set of setsAand
B asthe set ofelementswhich belong toAbut do not
belong to B. We denote the difference ofAand B by
A –B or simply“AminusB”.
A – B = {x : x Î A and x Ï B}.
A B

41
Sets
A B
1
2
6
7
4
5
3
A B
1
2
4
5
3
A – B = {1, 2, 3}
6
7
B – A = {6, 7}
m m
Example-6. Let A= {1, 2, 3, 4, 5}; B = {4, 5, 6, 7}. Find A– B.
Solution : GivenA= {1, 2, 3, 4, 5} and B = {4, 5, 6, 7}. Onlythe elements which are inAbut
not in B should be taken.
A – B = {1, 2, 3, 4, 5} – {4, 5, 6, 7} = {1, 2, 3}
A– B = {1, 2, 3}. Since 4, 5 are the elements in B theyare taken awayfromA.
Similarlyfor B–A, the elements which are onlyinB are taken.
B – A = {4, 5, 6, 7} – {1, 2, 3, 4, 5} = {6, 7}
B –A= {6, 7} (4, 5 are the elements inAand so they are taken away from B).
Note that A – B ¹ B – A
A
The Venn diagram of A – B and B – A are shown below.
DO THIS
1. IfA= {1, 2, 3, 4 ,5} and B = {4, 5, 6, 7}, then findA– B and B –A. Are they equal?
2. If V = {a, e, i, o, u} and B = {a, i, k, u}, find V – B and B – V.
THINK - DISCUSS
The setsA– B, B –A andAÇ B are mutuallydisjoint sets. Use examples to observe
ifthisis true.
EXERCISE - 2.2
1. If A= {1, 2, 3, 4} and B = {1, 2, 3, 5, 6}, then findAÇ Band B ÇA.Are they equal?
2. If A= {0, 2, 4}, find AÇ f and AÇ A. Comment.

Class-XMathematics
42
3. IfA= {2, 4, 6, 8, 10} and B = {3, 6, 9, 12, 15}, find A– B and B – A.
4. IfAand B are two sets such thatAÌ B then what isAÈ B?
5. Let A= {x : x is a naturalnumber}, B = {x: x is an even naturalnumber}
C = {x : x is an odd naturalnumber} and D = {x: x is a prime number}
Find A Ç B, AÇ C, AÇ D, B Ç C, B Ç D and C Ç D.
6. IfA = {3, 6, 9, 12, 15, 18, 21}, B = {4, 8, 12, 16, 20},
C = {2, 4, 6, 8, 10, 12, 14, 16} and D = {5, 10, 15, 20}, find
(i) A– B (ii)A– C (iii)A– D (iv) B – A (v) C – A
(vi) D –A (vii) B – C (viii) B– D (ix) C – B (x) D – B
7. State whether eachof the following statementsis true or false. Justifyyour answers.
(i) {2, 3, 4, 5} and {3, 6} are disjoint sets.
(ii) {a, e, i, o, u} and {a, b, c, d} are disjoint sets.
(iii) {2, 6, 10, 14} and {3, 7, 11, 15} are disjoint sets.
(iv) {2, 6, 10} and {3, 7, 11} are disjoint sets.
2.7 EQUAL SETS
Consider thefollowing sets.
A ={Sachin, Dravid, Kohli}
B ={Dravid, Sachin, Dhoni}
C = {Kohli, Dravid, Sachin}
What do you observe in the above three setsA, B and C?Allthe players that are in A
are inC. Also, allthe players that arein Care inA. Thus,Aand Chavesameelements but some
elements ofAand B are different. So, the setsAand C are equalsets but setsAand B are not
equal.
Two setsAand C are said to be equal if every element inAbelongs to C (i.e.AÍ C)
and every element in C belongs to A(i.e.C Í A).
If Aand C are equal sets, then we writeA= C. Thus, we can also write that C Í Aand
A Í C ÛA =C. [Here Û is the symbolfor two wayimplication andisusuallyread as, ifand
only if(brieflywrittenas “iff”]

43
Sets
Example-7. If A= {p, q, r} and B = {q, p, r}, then check whetherA=B or not.
Solution : GivenA= {p, q, r} and B = {q, p, r}.
In the above sets, every element ofAis also an element ofB. A Í B.
Similarlyeveryelement ofB is also inA. B Í A.
Then fromthe above two relations, we can sayA=B.
Examples-8. If A= {1, 2, 3, ….} and N is the set of natural numbers, then check whetherA
and N are equal?
Solution : The elements are same in both the sets. Therefore,A Í N and N Í A.
A.
Therefore, bothAand N are the set of Naturalnumbers. Therefore the setsAand N are
equal sets i.e. A= N.
Example-9. Consider the sets A= {p, q, r, s} and B = {1, 2, 3, 4}. Are they equal?
Solution :Aand B do not contain the same elements. So, A ¹ B.
Example-10. Let A be the set of prime numbers smaller than 6 and B be the set of prime
factors of 30. Check ifAand B are equal.
Solution : The set ofprime numbers less than 6, A= { 2,3,5}
The prime factors of 30 are 2, 3 and 5. So, P = { 2,3,5}
SincetheelementsofAare thesameas the elementsofPandvice versatherefore,Aand
P are equal. i.e ,
A B B A A B
Í Í Þ =
Example-11. Show that the sets C and B are equal, where
C = {x : x is a letter in the word ‘ASSASSINATION’}
B = {x : x is a letter in the word STATION}
Solution : Given that C = {x : x is a letter in the word ‘ASSASSINATION’}
The roster formofthe set C = {C,S,I,N,T,O}, since elementsin a set cannot be repeated.
Also given that B = {x : x is a letter in the word STATION}
‘B’can also be written asB = {A,S,I,N,T,O}
So, the elements of C and B are same and C = B.
i.e. C B, B C C=B
Í Í Þ

Class-XMathematics
44
Example-12. Consider the sets f , A = {1, 3}, B = {1, 5, 9}, C = {1, 3, 5, 7, 9}. Insert the
symbolÌ or Ë between each ofthe following pair ofsets.
(i) f ….. B (ii) A….. B (iii)A….. C (iv) B ….. C
Solution : (i) f Ì B, as f is a subset of everyset.
(ii) A Ë B, as 3 ÎAbut 3 Ï B.
(iii) A Ì C as 1, 3 ÎA also belong to C.
(iv) B Ì C as every element ofB is also anelement ofC.
EXERCISE - 2.3
1. Which ofthe following sets are equal?
A = {x : x is a letter in the word FOLLOW}, B = {x : x isa letter in theword FLOW}
and C = {x : x is a letter in the word WOLF}
2. Considerthe followingsetsand fillupthe blankswith= or ¹ so asto makethestatement
true.
A = {1, 2, 3}; B = {Thefirst three naturalnumbers}
C = {a, b, c, d}; D = {d, c, a, b}
E = {a, e, i, o, u}; F = {set ofvowels inEnglishAlphabet}
(i) A .... B (ii) A .... E (iii) C .... D
(iv) D .... F (v) F ....A (vi) D .... E
(vii) F .... B
3. In eachof the following, state whetherA= Bor not.
(i) A = {a, b, c, d} B = {d, c, a, b}
(ii) A = {4, 8, 12, 16} B = {8, 4, 16, 18}

45
Sets
(iii) A = {2, 4, 6, 8, 10} B= {x : x is a positive eveninteger andx < 10}
(iv) A = {x : x is a multiple of 10} B = {10, 15, 20, 25, 30, …}
4. State the reasonsfor the following :
(i) {1, 2, 3, …., 10} ¹ {x : x Î N and 1 < x < 10}
(ii) {2, 4, 6, 8, 10} ¹ {x : x = 2n+1 and x Î N}
(iii) {5, 15, 30, 45} ¹ {x : x is a multiple of 15}
(iv) {2, 3, 5, 7, 9} ¹ {x : x is a prime number}
5. List allthesubsetsofthefollowingsets.
(i) B = {p, q} (ii) C = {x, y, z} (iii) D = {a, b, c, d}
(iv) E = {1, 4, 9, 16} (v) F = {10, 100, 1000}
2.8 FINITE AND INFINITE SETS
Now considerthe following sets:
(i) A = {the students ofyour school} (ii) L= {p,q,r,s}
(iii) B = {x : x is an even number} (iv) J = {x : x is a multiple of 7}
Canyou list the number of elements in each ofthe sets givenabove? In(i), the number of
elementswillbe the number ofstudentsin your school. In(ii), thenumber ofelementsinset Lis
4. We find that it is possible to express the number ofelementsofsetsAand Lin definite whole
numbers. Such setsare called finite sets.
Now, consider the set B ofalleven numbers. We can not express the number of elements
in wholenumber i.e., we seethat the number ofelementsofthis set isnot finite.We find that the
number ofelements inB and J is infinite. Suchsets are called infinite sets.
We candraw infinite number ofstraight lines passing thougha given point. So this set is
infinite. Similarly, itisnot possible tofind out the last number among the collectionofallintegers.
Thus, we cansaya set is infinite ifit is not finite.
Consider somemore examples :
(i) Let ‘W’ be the set ofthe days ofthe week. ThenW is finite.

Class-XMathematics
46
(ii) Let ‘S’ be the set ofsolutions ofthe equation x2
– 16 = 0. Then S is finite.
(iii) Let ‘G’ be the set ofpoints ona line. Then G is infinite.
Example-13. State whichof the following sets are finite or infinite.
(i) {x : x Î ¥ and (x - 1) (x - 2) = 0} (ii) {x : x Î ¥ and x2
= 4}
(iii) {x : x Î ¥ and 2x - 2 = 0} (iv) {x : x Î ¥ and x is prime}
(v) {x : x Î ¥ and x is odd}
Solution :
(i) x can take the values 1 or 2 in the given case. The set is {1,2}. Hence, it is finite.
(ii) x2
= 4, implies that x = +2 or -2. But x Î ¥ or x is a natural number so the set
is{2}. Hence, it is finite.
(iii) In a given set x = 1 and 1 Î ¥ . Hence, it is finite.
(iv) The given set is the set of all prime numbers. There are infinitely many prime
numbers. Hence, set is infinite.
(v) Since thereare infinitenumber ofodd numbers, hence theset is infinite.
2.9 CARDINALITY OF A FINITE SET
Now, consider the following finite sets :
A = {1, 2, 4}; B = {6, 7, 8, 9, 10}; C = {x : x is a alphabet in the word "INDIA"}
Here,
Number ofelements in setA= 3.
Number ofelements in set B = 5.
Number ofelements in set C =4 (In the set C, the element ‘I’repeats twice. We know
that the elements of agiven set should be distinct. So, the number ofdistinct elements in set C
is 4).
The number of elements in a finite set is called the cardinal number of the set or the
cardinalityoftheset.
The cardinalnumber or cardinalityofthe setAis denoted as n(A) = 3.

47
Sets
Similarly, n(B) = 5 and n(C) = 4.
For finite set cardinalityis a whole number.
We willlearncardinalityofinfinite setsinhigher classes.
Note : There are no elements in a null set. The cardinal number of that set is n(f ) = 0
DO THESE
1. Which of the following are empty sets? Justify your answer.
(i) Set of integers which lie between 2 and 3.
(ii) Set of natural numbers that are smaller than 1.
(iii) Set of odd numbers that leave remainder zero, when divided by 2.
2. State which of the following sets are finite and which are infinite. Give reasons
for your answer.
(i) A = {x : x Î N and x < 100} (ii) B = {x : x Î N and x < 5}
(iii) C = {12
, 22
, 32
, …..} (iv) D = {1, 2, 3, 4}
(v) {x : x is a day of the week}.
3. Tick the set which is infinite
(A) The set of whole numbers < 10 (B) The set of prime numbers < 10
(C) The set of integers < 10 (D) The set of factors of 10
TRY THIS
1. Which of the following sets are empty sets? Justify your answer.
(i) A = {x : x2
= 4 and 3x = 9}.
(ii) The set of all triangles in a plane having the sum of their three angles less
than 180.

Class-XMathematics
48
THINK & DISCUSS
An empty set is a finite set. Is this statement true or false? Why?
EXERCISE - 2.4
1. State whichofthe following setsare emptyand whichare not?
(i) The set oflinespassingthrougha givenpoint.
(ii) Set ofodd naturalnumbers divisible by2.
(iii) {x : x is a natural number, x < 5 and x > 7}
(iv) {x : x is a common point to anytwo parallellines}
(v) Set ofeven prime numbers.
2. State whether the following sets are finite or infinite.
(i) The set ofmonths in a year. (ii) {1, 2, 3, …, 99, 100}
(iii) The set ofprime numbers smaller than 99.
(iv) The set ofletters intheEnglishalphabet.
(v) The set of lines that canbe drawn are parallelto the X-Axis.
(vi) The set ofnumbers whichare multiplies of5.
(vii) The set ofcircles passing through the origin(0, 0).
Example-14. If A= {1, 2, 3,4,5}; B = {2,4,6,8} then find n(AÈB).
Solution : The set Acontains five elements n(A) = 5
and the set B contains four elements n(B) = 4
ButAÈB={1,2,3,4,5,6,8} does not contain 9elements and it contains7 elements
only. Why?

49
Sets
THINK & DISCUSS
1. What is the relationbetween n(A), n(B), n(A Ç B) and n(AÈB)?
2. If Aand B are disjoint sets, then how canyou find n(AÈB)?
1. A set is a welldefined collectionof distinct objects, welldefined means that:
(i) There is a universe of objects which are allowed into consideration.
(ii) Anyobject in the universe is either an element or is not an element ofthe set.
2. An object belonging to a set is known asan element ofthe set. We use the symbol'Î'to
denote membershipofan element and read belongs to.
3. Sets canbe written in theroster formwhere allelements ofthe set are written, separated
bycommas, withincurlybrackets(braces).
4. Sets canalso be written in the set-builder form.
5. Aset whichdoesnot contain anyelement iscalled an empty set, or a Nullset, or a void
set.
6. Aset is called a finite set if its cordinality/ cardinalnumber is a whole number.
Suggested Projects
l Conduct a surveyin your classroomabout anytwo interested games/ newspapers/ TV
channels etc. andcollect the information.
Byusing sets, find out
(i) How manyare interested in game1/ newspaper1/ TV channel1?
(ii) How manyare interested in game2/ newspaper2/ TV channel2?
(iii) How manyareinterested in both? and
(iv) How manyareinterested inneither?
Extension: We can extend the above surveyfor three interested games/ newspapers/
TV channels etc.

Class-XMathematics
50
7. We cansaythat aset isinfinite ifit is not finite.
8. The number ofelements in afinite set is called the cardinalnumber/ cordinalityofthe set.
9. The universal set is denoted by 'm' or U. The universal set is usually represented by
rectangles.
10. Ais a subset ofB if 'a' is an element ofAimplies that 'a' is also an element ofB. This is
written asA Í B if a ÎA Þ a Î B, where A, B are two sets.
11. Two sets, Aand B are said to be equal if every element inAbelongs to B and every
element in B belongs toA.
12. Aunion B is written asA È B = {x : x ÎAor x Î B}.
13. Aintersection B is written asAÇ B = {x : x ÎAand x Î B}
14. The difference of two setsA, B is defined asA- B
A - B = {x : x Î A and x Ï B}
15. Venn diagrams are a convenient wayofshowing operations between sets.

3.1 INTRODUCTION
Let usobserve two situations.
1. A flower bed in a garden is in the shape of a triangle. The longest side is 3 times the
smallest side and the smallest side is 2 units shorter than the intermediate side. Let P
represents the lengthof the smallest side. Then what isthe perimeter interms of P?
2. The lengthofa rectangular dininghallis twice itsbreadth. Let x representsthe breadth of
the hall. What is the area ofthe floor ofthe hall in terms of x?
In the above situations, there is an"unknown" in each.In the first situation, the smallest
side isgiven as ‘P’units.
Since, Perimeter of triangle = sumofall sides
Perimeter = P + 3P + P + 2
= 5P + 2
Similarly inthe second situation, lengthis given as twicethe breadth.
So, if breadth = x, then length = 2x
Since area ofrectangle = lb
Area = (2x) (x)
= 2 x2
As you know, the perimeter, 5P+ 2 ofthe triangle and the area 2x2
ofthe rectangle are
intheformofpolynomials ofdifferent degrees.
2x
2 x2 x
Polynomials
3
3P
P+2
P

Class-X Mathematics
52
3.2 WHAT ARE POLYNOMIALS?
Apolynomialinxis anexpressioncontaining the sumofa finite number ofterms ofthe
form axn
for a real number a, where a ¹ 0 and a whole number n.
Polynomials Not polynomials
2x
1
2
4x
1
4
3
x - 3x2
+ 4x-1
+ 5
x2
- 2x - 1
1
4
x
+
Whyis
1
1
-
y
not a polynomial? Discuss with your friends and teacher.
.
DO THIS
State whichofthe following arepolynomialsand whicharenot? Give reasons.
(i) 2x3
(ii) ∋ (
1
1 0
x
x
, ÷ (iii)
2 1
4
7
z ∗ (iv) 2
2 2
m m
, ∗ (v) 2
1
P,
∗
3.2.1 DEGREE OF A POLYNOMIAL
Recallthat ifp(x) is apolynomialin x, the highest power ofxinp(x) iscalled the degree
ofthe polynomialp(x). Forexample, 3x + 5 is a polynomialinthe variable x. It is ofdegree 1 and
is called a linear polynomial. 5x, 2 5
y∗ ,
1
3
P,
, m + 1 etc. are some more linear polynomials.
A polynomial of degree 2 is called a quadratic polynomial. For example, x2
+ 5x + 4 is
a quadratic polynomialin the variable x. 2x2
+ 3x -
1
2
, p2
-1, 3 – z – z2
, y2
-
3
y
+ 2 are some
examples ofquadratic polynomials.
The expression 5x3
–4x2
+x–1 is a polynomialin the variable x ofdegree 3, and is called
a cubic polynomial. Some more examples of cubic polynomials are 2 – x3
, p3
, l3
– l2
– l + 5.
6 can be written as 6´ x0
.As the index of x is 0, it is a polynomial of 0 degree.
TRY THIS
Write3different quadratic,cubicand 2linear polynomialswithdifferentnumber ofterms.
We can write polynomials of any degree. 7 u 6
–
3
2
u 4
+ 4 u 2
– 8 is polynomial of
degree 6 and x10
– 3x8
+ 4x5
+ 2x2
-1 is a polynomial of degree 10.

Polynomials 53
We can write a polynomial in a variable x ofa degree n where n is anywhole number.
Generally, we say
p(x) = a0
xn
+ a1
xn-1
+ a2
xn-2
+ …….. + an-1
x +an
is a polynomial of nth
degree,
where a0
, a1
, a2
….. an-1,
an
are real coefficients of x and a0
¹ 0
For example, the general form of a first degree polynomial in one variable x is ax+b,
where a and b are real numbers and a ¹ 0.
TRY THIS
1. Write the generalformofa quadratic polynomialand a cubic polynomialin variable x.
2. Write a generalpolynomialq(z) ofdegree n with coefficients that are b0, b1, b2, ..... bn.
What are the conditions on b0, b1, b2, ..... bn?
3.2.2 VALUE OF A POLYNOMIAL
Now consider the polynomialp(x) = x2
– 2x – 3. What is the value ofthe polynomialat
anyvalue of x? For example, what is the value at x = 1? Substituting x = 1, inthe polynomial,
we get p(1) = (1)2
– 2(1) – 3 = –4. The value – 4, is obtained byreplacing x by 1 in the given
polynomial p(x). This is the value of x2
– 2x – 3 at x = 1.
Similarly, p(0) = –3 is the value of p(x) at x = 0.
Thus, if p(x) is a polynomial in x, and if k is a real number, then the value obtained by
replacing x by k in p(x), is called the value of p(x) at x = k, and is denoted by p(k).
DO THIS
(i) Ifp(x)=x2
–5x–6, thenfind thevaluesof p(1), p(2), p(3), p(0), p(–1), p(–2),p(–3).
(ii) If p(m) = m2
– 3m + 1, thenfind the value of p(1) and p(–1).
3.2.3 ZEROES OF A POLYNOMIAL
What arevalues of p(x) = x2
– 2x – 3 at x = 3, –1 and 2?
We have, p(3) = (3)2
– 2(3) – 3 = 9 – 6 – 3 = 0
Also p(-1) = (–1)2
– 2(–1) – 3 = 1 + 2 – 3 = 0
and p(2) = (2)2
– 2(2) – 3 = 4 – 4 – 3 = –3

Class-X Mathematics
54
We see that p(3) = 0 and p(-1) = 0. 3 and –1 are called Zeroes of the polynomial
p(x) = x2
– 2x -3.
As p(2) ¹ 0, 2 is not zero of p(x).
More generally, a real number k is said to be a zero ofa polynomial p(x), if p(k) = 0.
DO THIS
(i) Let p(x) = x2
– 4x + 3. Findthe value of p(0), p(1), p(2), p(3) and obtainzeroes of the
polynomialp(x).
(ii) Check whether -3 and 3 are the zeroes ofthe polynomial x2
– 9.
EXERCISE - 3.1
1. In p(x) = 5x7
– 6x5
+ 7x-6, what is the
(i) coefficient ofx5
(ii) degree ofp(x) (iii) constant term.
2. State which ofthe following statements are true and which are false? Give reasons for
your choice.
(i) The degree ofthe polynomial 2 x2
– 3x + 1 is 2.
(ii) The coefficient of x2
in the polynomial p(x) = 3x3
– 4x2
+ 5x + 7 is 2.
(iii) The degreeof a constant termis zero.
(iv) 2
1
5 6
x x
, ∗
is a quadratic polynomial.
(v) The degree ofa polynomialis one more than the number ofterms in it.
3. If p(t) = t3
– 1, find the values of p(1), p(–1), p(0), p(2), p(–2).
4. Check whether –2 and 2 are the zeroes ofthe polynomial x4
– 16.
5. Check whether 3 and –2 are the zeroes of the polynomial p(x) when p(x) = x2
– x – 6.

Polynomials 55
3.3 WORKING WITH POLYNOMIALS
You have alreadylearned how to find the zeroesofa linear polynomial.
For example, if k is a zero of p(x) = 2x + 5, then p(k) =0 gives 2k +5 = 0 k =
5
2
,
.
In general, if k is a zero of p(x) = ax+b (a ¹ 0), then p(k) = ak + b = 0,
Therefore k =
b
a
,
, or the zero of the linear polynomial ax + b is
b
a
,
.
Thus, the zero ofa linear polynomialis related to its coefficients,including the constant
term.
Are the zeroes of higher degree polynomials also related to their coefficients? Think
about this and discuss with friends. We willcome to this later.
3.4 GEOMETRICAL MEANING OF THE ZEROES OF A POLYNOMIAL
You know that a real number k is a zero of the polynomial p(x) if p(k) = 0. Let us see the
graphical representations oflinear and quadratic polynomials and thegeometricalmeaning of
theirzeroes.
3.4.1. GRAPHICAL REPRESENTATION OF A LINEAR POLYNOMIAL
Consider first alinear polynomial ax + b (a ¹ 0).You have studiedin Class-IX that the graph of
y = ax + b is a straight line. For example, the graph of y = 2x + 3 is a straight line intersecting
the Y-axis at (0, 3) and it also passes throughthe points (–2, –1) and (2, 7).
Table 3.1
x –2 –1 0 2
y = 2x + 3 –1 1 3 7
(x, y) (–2, –1) (–1, 1) (0, 3) (2, 7)

Class-X Mathematics
56
1
2
3
4
5
6
7
-1
-1
-2
-2
-3
-3
-4
-4
-5
-6
-7
0
1 2 3 4 5 6 7
-2
-3
-4
-5
-6
X' X
-7
Y
Y'
(-2, -1)
3
,0
2
-
æ ö
ç ÷
è ø
(0, 3)
(2, 7)
In the graph, you can see
that the graph of y = 2x+3
intersects the X-axis between x
= –1 and x = –2, that is,
at the point
3
, 0
2
-
æ ö
ç ÷
è ø
. But
x =
3
2
-
is the zero of the
polynomial 2x + 3. Thus, the
zero of the polynomial
2x + 3 is the x-coordinate ofthe
point where the graph of y
= 2x + 3 intersects the
X-axis.
DO THIS
Draw the graph of (i) y = 2x + 5, (ii) y = 2x – 5, (iii) y = 2x and find the point of
intersection on X-axis. Is the x-coordinate ofthese points also the zeroesofthe polynomial?
In general, for alinear polynomialax + b, a ¹ 0, the graphof y = ax + b is a straight line
whichintersects the X-axis atexactlyone point, namely, ( ,0)
b
a
,
.
Therefore, the linear polynomial ax + b, a ¹ 0, has exactly one zero, namely, the
x-coordinate of the point where the graph of y = ax + b intersects the X-axis.
3.4.2. GRAPHICAL REPRESENTATION OF A QUADRATIC POLYNOMIAL
Now,letuslookfor thegeometricalmeaningofazero ofaquadraticpolynomial.Consider
the quadratic polynomialx2
– 3x – 4. Let us see how the graph of y = x2
– 3x – 4 looks like. Let
us list a few values of y = x2
– 3x – 4 corresponding to a few values for x as giveninTable 3.2.
Table 3.2
x – 2 – 1 0 1 2 3 4 5
y = x2
– 3x – 4 6 0 – 4 – 6 – 6 – 4 0 6
(x, y) (– 2, 6) (– 1, 0) (0, -4) (1, – 6) (2, – 6) (3, – 4) (4, 0) (5, 6)

Polynomials 57
We locate thepoints listed above
on a graph paper and draw the
graph.
Is the graph of y = x2
– 3x – 4 a
straight line? No, it is like a
shaped curve. It is intersecting
the X-axis at two points.
In fact, for any quadratic
polynomial ax2
+ bx + c,
a ¹ 0, the graph of the
corresponding equation
y = ax2
+ bx + c ( )
0
a ¹ either
opens upwardslike or opens
downwards like . This
depends on whether a > 0 or a
< 0. (The shape of these curves
are called parabolas.)
Fromthe table, we observe that -1 and 4 are zeroes ofthe quadratic polynomial. From
the graph, we see that -1 and 4 arealso X coordinates ofpoints ofintersection ofthe parabola
with the X-axis. Zeroes of the quadratic polynomial x2
– 3x – 4 are the x-coordinates of the
points where the graph of y = x2
– 3x – 4 intersects the X-axis. For the polynomial P(x) = y =
x2
– 3x – 4; P(-1)=0, its graph is intersecting the X-axis at (-1, 0). Also P(4)=0 its graph is
intersecting the X-axis at (4, 0). In general for polynomialP(x) if P(a)=0, its graph intersects
X-axis at (a, 0).
This is true for any quadratic polynomial, i.e., the zeroes of a quadratic polynomial
ax2
+ bx + c, ( )
0
a ¹ are precisely the x-coordinates of the points where the parabola
representing y = ax2
+ bx + c ( )
0
a ¹ intersects the X-axis.
TRY THIS
Draw the graphs of (i) y = x2
– x – 6 (ii) y = 6– x – x2
and find zeroes in each case.
What do you notice?
1
2
3
4
5
6
7
-1
-1
-2
-2
-3
-3
-4
-4
-5
-6
-7
0
1 2 3 4 5 6 7
-2
-3
-4
-5
-6
X' X
-7
Y
Y'
(-2, 6)
(-1, 0)
(0, -4)
(1, -6)
(2, -6)
(3, -4)
(4, 0)
(5, 6)

Class-X Mathematics
58
Fromour earlier observation about the shape ofthe graph of y= ax2
+ bx + c, ( )
0
a ¹
the following three cases arise.
Case (i) : Here, the graph cuts X-axis at two distinct points A and A¢ . In this case, the
x-coordinates ofA and A¢ are the two zeroes of the quadratic polynomial ax2
+bx +c . The
parabola opens either upward or downward.
(i) (ii)
Case (ii) : Here, thegraphtouches X-axis at exactlyone point, i.e., at two coincident points. So,
the two pointsAand A¢ ofCase (i) coincide here to become one pointA.
(i) (ii)
In this case, the x-coordinate ofAis the only zero for the quadratic polynomial ax2
+ bx + c.
X' X
Y
Y'
X' X
Y
Y'
A
A'
O A' A
O
X' X
Y
Y'
X' X
Y
Y'
A
O A
O

Polynomials 59
Case (iii) : Here, the graph is either completely above the X-axis or completely below the
X-axis. So, it does not cut the X-axis at any point.
(i) (ii)
So, the quadratic polynomial ax2
+ bx + c has no zero inthis case.
So, you can see geometricallythat a quadratic polynomial canhave either two distinct zeroes or
two equalzeroes (i.e., one zero), or no zero. This also means that a polynomialofdegree 2 has
atmost two zeroes.
TRY THIS
1. Write three quadratic polynomialsthat have 2 zeroes each.
2. Write onequadratic polynomialthat has one zero.
3. How willyou verifyifaquadratic polynomialhas onlyone zero?
4. Write three quadraticpolynomials that haveno zeroes.
3.4.3 GEOMETRICAL MEANING OF ZEROES OF A CUBIC POLYNOMIAL
What do you expect the geometricalmeaning of the zeroesofa cubic polynomialto be?
Let us find out. Consider the cubic polynomial x3
– 4x. To see how the graph of y = x3
– 4x
looks like, let us list a few values of y corresponding to a few values for x asshown inTable 3.3.
X' X
Y
Y'
O X' X
Y
Y'
O

Class-X Mathematics
60
Table 3.3
x –2 –1 0 1 2
y = x3
– 4x 0 3 0 –3 0
(x, y) (–2, 0) (–1, 3) (0, 0) (1, –3) (2, 0)
We see that the graph of y
= x3
– 4x looks like the one
giveninthe figure.
We see from the table above
that –2, 0 and 2 are zeroes of
the cubic polynomial x3
– 4x.
–2, 0 and 2 are the
x-coordinates of the points
where the graph of y = x3
–
4x intersects the X-axis. So
thispolynomialhasthreezeros.
Let us take a few more
examples. Consider the cubic
polynomials x3
and x3
– x2
respectively. See Table 3.4
and 3.5.
Table 3.4
x –2 –1 0 1 2
y = x3
–8 –1 0 1 8
(x, y) (–2, –8) (–1, –1) (0, 0) (1, 1) (2, 8)
1
2
3
4
5
6
7
-1
-1
-2
-2
-3
-3
-4
-4
-5
-6
-7
0
1 2 3 4 5 6 7
-2
-3
-4
-5
-6
X' X
-7
Y
Y'
(1, -3)
(2, 0)
(-1, 3)
(-2, 0)
Scale: X-axis: 1cm= 1 unit.
Y axis 1 cm= 1 unit.

Polynomials 61
-1
-1
-2
-2
-3
-3
-4
-4
-5
-6
-7
-2
Y'
1
2
3
4
5
6
7
-1
-1
-2
-2
-3
-3
-4
-4
-5
-6
-7
0
1 2 3 4
-2
-3
-4
X' X
Y
Y'
(-1, -2)
(1, 0)
(2, 4)
(0, 0)
Table 3.5
x –2 –1 0 1 2
y = x3
– x2
–12 –2 0 0 4
(x, y) (–2, –12) (–1, –2) (0, 0) (1, 0) (2, 4)
y = x3
y = x3
– x2
In y = x3
, you can see that 0 is the x-coordinate of the only point where the graph of
y = x3
intersects the X-axis. So, the polynomialhasonlyone distinct zero. Similarly, 0 and1 are
the x-coordinates ofthe onlypoints where thegraph ofy = x3
– x2
intersects the X-axis. So, the
cubic polynomialhastwo distinct zeroes.
Fromtheexamplesabove, weseethat thereareatmost 3zeroesforanycubicpolynomial.
In other words, any polynomialofdegree 3 can have at most three zeroes.
TRY THIS
Find the zeroesofcubic polynomials (i) – x3
(ii) x2
–x3
(iii) x3
– 5x2
+ 6x without drawing
the graph of the polynomial.
1
2
3
4
5
6
7
-1
-1
-2
-2
-3
-3
-4
-4
-5
-6
-7
0
1 2 3 4
-2
-3
-4
X' X
Y
Y'
(-2, -8)
(-1, -1)
(2, 8)
(1, 1)
(0, 0)
Scale:
X=axis:1cm=1 unit
Y=axis:1cm=1 unit
Scale:
X=axis:1cm=1 unit
Y=axis:1cm=1 unit

Class-X Mathematics
62
Remark : In general, given a polynomial p(x) of degree n, the graph of y = p(x) intersects the
X-axis at at most n points. Therefore, a polynomialp(x) of degree n has at most n zeroes.
Example-1. Look at the graphs in the figuresgiven below. Each is the graphof y = p(x),where
p(x) is a polynomial. In each ofthe graphs, find the number ofzeroes of p(x) in thegiven range
of x.
Solution : In the given range of x inrespective graphs :
(i) The number ofzeroes is 1 as the graphintersects the X-axisat one point only.
(ii) The number of zeroes is 2 as the graph intersects the X-axis at two points.
(iii) The number ofzeroes is 3. (Why?)
(iv) The number ofzeroes is 1. (Why?)
(v) The number ofzeroes is 1. (Why?)
(vi) The number ofzeroes is 4. (Why?)
(i) (ii) (iii)
(iv) (v) (vi)
X' X
Y
Y'
X' X
Y
Y'
X' X
Y
Y'
O O
O
X' X
Y
Y'
X' X
Y
Y'
X' X
Y
Y'
O O
O

Polynomials 63
Example-2. Find thenumber of zeroes ofthe given polynomials.And also find their values.
(i) p(x) = 2x + 1 (ii) q(y) = y2
– 1 (iii) r(z)= z3
Solution :We willdo thiswithout plotting the graph.
(i) p(x) = 2x + 1 is a linear polynomial. It has only one zero.
To find zeroes,
Let p(x) = 0
So, 2x+1=0
Therefore x =
1
2
,
The zero ofthe given polynomialis
1
2
,
.
(ii) q(y) = y2
– 1 is a quadratic polynomial.
It has at most two zeroes.
To find zeroes, let q(y) = 0
Þ y2
– 1 = 0
Þ (y + 1) (y – 1) = 0
Þ y = -1 or y = 1
Therefore the zeroes ofthe polynomialare -1 and 1.
(iii) r(z) = z3
is a cubic polynomial. It has at most three zeroes.
Let r(z) = 0
Þ z3
= 0
Þ z = 0
So, thezero ofthe polynomialis 0.

Class-X Mathematics
64
EXERCISE – 3.2
1. The graphs ofy = p(x) are giveninthe figure below, for some polynomials p(x).Ineach
case, find the number ofzeroes of p(x).
2. Find the zeroes ofthe givenpolynomials.
(i) p(x) = 3x (ii) p(x) = x2
+ 5x + 6
(iii) p(x) = (x+2) (x+3) (iv) p(x) = x4
– 16
3. Draw the graphs of the given polynomialand find the zeroes. Justifythe answers.
(i) p(x) = x2
– x – 12 (ii) p(x) = x2
– 6x + 9
(iii) p(x) = x2
– 4x + 5 (iv) p(x) = x2
+ 3x – 4
(v) p(x) = x2
– 1
4. Why are
1
4
and –1 zeroes of the polynomials p(x) = 4x2
+ 3x – 1?
(i) (ii) (iii)
(iv) (v) (vi)
X' X
Y
Y'
X' X
Y
Y'
X' X
Y
Y'
O O
O
X' X
Y
Y'
X' X
Y
Y'
X' X
Y
Y'
O O
O

Polynomials 65
3.5 RELATIONSHIP BETWEEN ZEROES AND COEFFICIENTS OF A POLYNOMIAL
You have alreadyseenthat the zero ofa linear polynomialax + b is–
b
a
.Wewillnow try
to explore the relationship between zeroes and coefficients ofa quadraticpolynomial. For this,
let us take the quadratic polynomial p(x) = 2x2
– 8x + 6.
In Class-IX, we have learnt how to factorise quadratic polynomials by splitting the
middle term. So, here wesplit the middle term‘–8x’ as a sumoftwo terms, whoseproduct is
6 × 2x2
= 12 x2
. So, we write
2x2
– 8x + 6 = 2x2
– 6x – 2x + 6
= 2x(x – 3) – 2(x – 3)
= (2x – 2) (x – 3) = 2(x – 1) (x – 3)
p(x) = 2x2
– 8x + 6 is zero when x – 1 = 0 or x – 3 = 0, i.e., when x = 1 or x = 3. So,
the zeroes of2x2
–8x + 6are 1 and 3.We now tryand see ifthese zeroeshave some relationship
to the coefficients of terms in the polynomial. The coefficient of x2
is 2; of xis –8 and the
constant is 6, whichis the coefficient of x0
. (i.e. 6x0
= 6)
We see that the sum of the zeroes = 1 + 3 = 4 =
( 8)
2
, ,
= 2
(coefficient of )
coefficientof
x
x
,
Product ofthe zeroes = 1 × 3 = 3 =
6
2
= 2
constant term
coefficient of x
Let us takeone more quadratic polynomial:
p(x) = 3x2
+ 5x – 2.
Bysplitting the middle termwe see,
3x2
+ 5x – 2 = 3x2
+ 6x – x – 2 = 3x(x + 2) – 1(x + 2)
= (3x – 1) (x + 2)
3x2
+ 5x – 2 is zero when either 3x – 1 = 0 or x + 2 = 0
i.e., when x =
1
3
or x = –2.
The zeroes of 3x2
+ 5x – 2 are
1
3
and –2. We can see that the :
Sum ofits zeroes =
1
3
+ (-2) =
5
3
,
= 2
(coefficient of )
coefficientof
x
x
,
Product ofits zeroes =
1
3
× (-2) =
2
3
,
= 2
constant term
coefficient of x

Class-X Mathematics
66
DO THIS
Find the zeroesofthe quadratic polynomials givenbelow. Find the sumand product
ofthe zeroesand verifyrelationship to the coefficients oftermsin the polynomial.
(i) p(x) = x2
– x – 6 (ii) p(x) = x2
– 4x + 3
(iii) p(x) = x2
– 4 (iv) p(x) = x2
+ 2x + 1
In general, if a and b are the zeroes of the quadratic polynomial p(x) = ax2
+ bx + c,
where a ¹ 0, then (x – a) and (x – b) are the factors of p(x). Therefore,
ax2
+ bx + c = k (x – a) (x – b), where k is a constant
= k[x2
– (a+b) x + ab]
= k x2
– k (a+b) x + kab
Comparing the coefficients ofx2
, x and constant terms on both the sides, we get
a = k, b = – k(a+b) and c = kab.
This gives a + b =
b
a
,
,
ab =
c
a
Note : a and b are Greek letters pronounced as ‘alpha’and ‘beta’respectively. We will
use onemore letter ‘g’ pronounced as ‘gamma’.
Sum ofzeroes for a quadratic polynomial ax2
+ bx + c
= a + b =
b
a
,
= 2
(coefficient of )
coefficientof
x
x
,
Product of zeroes for a quadratic polynomial ax2
+ bx + c
= ab =
c
a
= 2
constant term
coefficient of x
Let us consider some examples.
Example-3. Find the zeroesofthe quadratic polynomialx2
+ 7x+10, and verifythe relationship
betweenthe zeroes and the coefficients.
Solution : We have
x2
+ 7x + 10 = (x + 2) (x + 5)

Polynomials 67
So, the value of x2
+ 7x + 10 is zero when x + 2 = 0 or x + 5 = 0,
i.e., when x = –2 or x = –5.
Therefore, the zeroes of x2
+ 7x + 10 are –2 and –5.
Now, sumofthe zeroes= –2 + (–5) = – (7) =
(7)
1
,
= 2
(coefficient of )
coefficientof
x
x
,
Product ofthe zeroes = –2 × (–5) = 10 =
10
1
= 2
constant term
coefficient of x
Example-4. Find the zeroes ofthe polynomial x2
– 3 and verify the relationship between the
zeroesandthe coefficients.
Solution : Recall the identity a2
– b2
= (a – b) (a + b).
Using it, we can write:
x2
– 3 = (x – 3 ) (x + 3 )
So, the value of x2
– 3 is zero when x = 3 or x = – 3 .
Therefore, the zeroes of x2
– 3 are 3 and – 3 .
Sum of the zeroes = 3 + (– 3 ) = 0 = 2
(coefficient of )
coefficientof
x
x
,
Product of zeroes = ( 3 ) × (– 3 ) = – 3 =
3
1
,
= 2
constant term
coefficient of x
Example-5. Find the quadratic polynomial, whose sumand product ofthe zeroes are – 3 and
2, respectively.
Solution : Let the quadratic polynomial be ax2
+ bx + c, and its zeroes be a and b. We have
a + b = – 3 =
b
a
,
,
and ab = 2 =
c
a
.
If we take a = 1, then b = 3 and c = 2
So, one quadratic polynomialwhichfits the given conditions is x2
+ 3x+ 2.

Class-X Mathematics
68
Similarly, we cantake 'a' to be anyreal number. Let us sayit is k. This gives 3
b
k
-
= -
or b = 3k and 2
c
k
= or c = 2k. Substituting the values of a, b and c, we get the polynomialis
kx2
+ 3kx + 2k.
Example-6. Find the quadratic polynomialwhose zeroes are 2 and
1
3
,
.
Solution : Let thequadratic polynomialbe
ax2
+ bx + c, a ¹ 0 and its zeroes be a and b.
Here a = 2, b =
1
3
,
Sum of the zeroes = (a + b) = 2 +
1
3
æ ö
, ÷
ç ÷
ç ÷
ç
è ø
=
5
3
Product ofthe zeroes = (ab) = 2
1
3
æ ö
, ÷
ç ÷
ç ÷
ç
è ø
=
2
3
,
Therefore the quadratic polynomial ax2
+ bx + c is
k[x2
– (a+b)x + a b], where k is a constant and k ¹ 0
i.e. k[x2
–
5
3
x –
2
3
]
We can take different values for k.
When k = 3, the quadratic polynomial willbe 3x2
– 5x – 2.
TRY THIS
(i) Find a quadratic polynomialwith zeroes -2 and
1
3
.
(ii) What is the quadratic polynomial the sum of whose zeroes is
3
2
-
and the product
of the zeroes is -1.

Polynomials 69
3.6 CUBIC POLYNOMIALS
Let us now look at cubic polynomials. Do you think similar relationholds between the
zeroes of acubic polynomialand its coefficients as well?
Let us consider p(x) = 2x3
– 5x2
– 14x + 8.
We see that p(x) = 0 for x = 4, – 2,
1
2
.
Since p(x) can have at most three zeroes, these are the zeroes of 2x3
– 5x2
– 14x + 8.
Sum ofits zeroes = 4 + (–2) +
1
2
=
5
2
=
( 5)
2
, ,
=
2
3
(coefficient of )
coefficientof
x
x
,
Product ofits zeroes = 4 × (–2) ×
1
2
= – 4 =
8
2
,
= 3
(constant term)
coefficientof x
,
However, there is one more relationship here. Consider the sumofthe products ofthe
zeroes takentwo at a time.We have:
=
1 1
{4 ( 2)} ( 2) 4
2 2
ì ü ì ü
ï ï ï ï
ï ï ï ï
≥ , ∗ , ≥ ∗ ≥
í ý í ý
ï ï ï ï
ï ï ï ï
î þ î þ
= – 8 – 1 + 2 = – 7 =
14
2
,
= 3
constant of
coefficient of
x
x
In general, it canbe proved that if a, b, g are the zeroes ofthe cubic polynomial
ax3
+ bx2
+ cx + d,
a + b + g =
b
a
,
,
ab + bg + ga =
c
a
and a b g =
d
a
,
.
3 2
ax bx cx d
+ + + is a polynomial with zeroes , ,
a b g . Let us
see how , ,
a b g relate to a, b, c, d.
Since , ,
a b g are the zeroes, the polynomialcanbe written as
( )( )( )
x x x
- a -b - g
3 2
( ) ( )
= - a + b + g + ab + bg + ag - abg
x x x
To compare with the polynomial, we multiplyby'a'andget
3 2
( ) ( )
ax x a xa a
- a +b + g + ab +bg + ag - abg .
( ), ( ),
= - a + b + g = ab + bg + ag = - abg
b a c a d a

Class-X Mathematics
70
DO THIS
Ifa, b, g are the zeroes of the givencubic polynomials, find the values of the expressions
giveninthetable.
S.No. Cubic Polynomial a + b + g ab + bg + ga ab g
1 x3
+ 3x2
– x – 2
2 4x3
+ 8x2
– 6x – 2
3 x3
+ 4x2
– 5x – 2
4 x3
+ 5x2
+ 4
Let us consider an example.
Example-7. Verifywhether 3, –1 and –
1
3
are the zeroes ofthe cubic polynomial
p(x) =3x3
–5x2
– 11x– 3,and thenverifytherelationshipbetweenthe zeroesand the coefficients.
Solution : p(x) = 3x3
– 5x2
– 11x – 3 is the given polynomial.
Then p(3) = 3 × 33
– (5 × 32
) – (11 × 3) – 3 = 81 – 45 – 33 – 3 = 0,
p(–1) = 3 × (–1)3
– 5 × (–1)2
– 11 × (–1) – 3 = – 3 – 5 + 11 – 3 = 0,
3 2
1 1 1 1
3 5 11 3
3 3 3 3
p
æ ö æ ö æ ö æ ö
÷ ÷ ÷ ÷
ç ç ç ç
, < ≥ , , ≥ , , ≥ , ,
÷ ÷ ÷ ÷
ç ç ç ç
÷ ÷ ÷ ÷
ç ç ç ç
è ø è ø è ø è ø
,
1 5 11 2 2
3 0
9 9 3 3 3
<, , ∗ , <, ∗ <
Therefore, 3, –1, and
1
3
, are the zeroes of 3x3
– 5x2
– 11
1x – 3.
So, we take a = 3, b = –1 and g =
1
3
, .
Comparing the given polynomial with ax3
+ bx2
+ cx + d, we get
a = 3, b = – 5, c = – 11, d = – 3.
Now,
a + b + g = 3 + (–1) +
1
3
æ ö
÷
ç, ÷
ç ÷
ç
è ø
= 2
1
3
, =
5
3
=
( 5)
3
, ,
=
b
a
,
,
ab + bg + ga = 3 × (–1) + (–1) ×
1
3
æ ö
÷
ç, ÷
ç ÷
ç
è ø
+
1
3
æ ö
÷
ç, ÷
ç ÷
ç
è ø
× 3 = – 3 +
1
3
– 1 =
11
3
,
=
c
a
,
a b g = 3 × (–1) ×
1
3
æ ö
÷
ç, ÷
ç ÷
ç
è ø
= 1 =
( 3)
3
, ,
=
d
a
,
.

Polynomials 71
EXERCISE – 3.3
1. Findthezeroesofthefollowingquadraticpolynomialsandverifytherelationshipbetween
the zeroesand the coefficients.
(i) x2
– 2x – 8 (ii) 4s2
– 4s + 1 (iii) 6x2
– 3 – 7x
(iv) 4u2
+ 8u (v) t2
– 15 (vi) 3x2
– x – 4
2. Find the quadratic polynomial in each case, with the given numbers as the sum and
product ofits zeroesrespectively.
(i)
1
4
, – 1 (ii)
1
2,
3
(iii) 0, 5
(iv) 1, 1 (v) –
1
4
,
1
4
(vi) 4, 1
3. Find the quadratic polynomial, for the zeroes a, b givenin each case.
(i) 2, –1 (ii) 3 , – 3 (iii)
1
4
, – 1 (iv)
1
2
,
3
2
4. Verify that 1, –1 and +3 are the zeroes of the cubic polynomial x3
– 3x2
– x + 3 and
check the relationship between zeroes andthe coefficients.
3.7 DIVISION ALGORITHM FOR POLYNOMIALS
You know that a cubic polynomialhas at most three zeroes. However, ifyou are given
only one zero, can you find the other two? For example, let us consider the cubic polynomial
x3
+ 3x2
– x – 3. If one of its zeroes is 1, then you know that this polynomial is divisible by
x – 1. Therefore dividing by x – 1 we would get the quotient x2
– 2x – 3.
We get thefactors of x2
– 2x – 3 bysplitting the middle term. The factorsare (x + 1) and
(x – 3). This gives us
x3
– 3x2
– x + 3 = (x – 1) (x2
– 2x – 3)
= (x – 1) (x + 1) (x – 3)
So, the three zeroes of thecubic polynomialare 1, – 1, 3.
Let usdiscussthe method ofdividing one polynomialbyanother indetail. Before doing
the steps formally, consider an example.

Class-X Mathematics
72
Example-8. Divide 2x2
+ 3x + 1 by x + 2.
Solution : Note that we stop the division process when either the
remainder is zero or its degree isless than the degree ofthe divisor. So,
here the quotient is 2x – 1 and the remainder is 3.
Let usverifydivisionalgorithm.
(2x – 1) (x + 2) + 3 = 2x2
+ 3x – 2 + 3 = 2x2
+ 3x + 1
i.e., 2x2
+ 3x + 1 = (x + 2) (2x – 1) + 3
Therefore, Dividend = Divisor × Quotient + Remainder
Let us now extend this process to divide a polynomial by a
quadratic polynomial.
+ x2
+ 2x + 5 by 1 + 2x + x2
.
Solution : We first arrange the terms of the dividend
and the divisor in the decreasing order of their degrees.
(Arranging the terms in this order is termed as writing
the polynomials in its standard form). In this example,
thedividendisalreadyinitsstandardform,andthedivisor
is also in standard form, is x2
+ 2x + 1.
Step 1 : To obtainthe first termofthe quotient, divide
the highest degree termofthe dividend (i.e., 3x3
)bythe
highest degree term of the divisor (i.e., x2
). This is 3x.
Then carryout the division process. What remains is –5x2
–x+5.
Step 2 : Now, to obtain the second termofthe quotient, divide the highest degree termofthe
new dividend (i.e., – 5x2
) bythehighest degree termofthe divisor (i.e., x2
). Thisgives–5.Again
carry out the division process with– 5x2
– x + 5.
Step 3 : What remains is 9x + 10. Now, the degree of 9x + 10 is less than the degree of the
divisor x2
+ 2x + 1. So, we cannot continue the division anyfurther.
So, the quotient is 3x – 5 and the remainder is 9x + 10.Also,
(x2
+ 2x + 1) × (3x – 5) + (9x + 10) = (3x3
+ 6x2
+ 3x – 5x2
– 10x – 5 + 9x + 10)
= 3x3
+ x2
+ 2x + 5
Here again, we see that
Dividend = Divisor × Quotient + Remainder
2
2
2 1
2 2 3 1
2 4
1
2
3
x
x x x
x x
x
x
,
∗ ∗ ∗
∗
, ,
, ∗
, ,
∗ ∗
2 3 2
3 2
2
2
3 5
2 1 3 2 5
3 6 3
5 5
5 10 5
9 10
x
x x x x x
x x x
x x
x x
x
,
∗ ∗ ∗ ∗ ∗
∗ ∗
, , ,
, , ∗
, , ,
∗ ∗ ∗
∗

Polynomials 73
We are applying here analgorithmcalledEuclid’sdivisionalgorithm.
Thissaysthat
Ifp(x) andg(x)areanytwopolynomialswith g(x)¹ 0, then wecan find polynomials
q(x) and r(x) such that
p(x) = g(x) × q(x) + r(x),
where either r(x) = 0 or degree of r(x) < degree of g(x) if r(x) ¹ 0
Thisresult is knownas the DivisionAlgorithmfor polynomials.
Now, we havethe following results fromthe above discussions
(i) If g(x) is alinear polynomial then r(x) = r is a constant.
(ii) If degree of g(x) = 1, then degree of p(x) = 1 + degree of q(x).
(iii) If p(x) is divided by(x – a), thenthe remainder is p(a).
(iv) If r = 0, we say q(x) divides p(x) exactly or q(x) is a factor of p(x).
Let usnow take some examples to illustrate its use.
– x3
– 3x + 5 by x – 1 – x2
, and verify the division algorithm.
Solution : Note that the given polynomials are not in standard form. To carryout division, we
first write both the dividend and divisor indecreasing orders oftheir degrees.
So, dividend = – x3
+ 3x2
– 3x + 5 and
divisor = – x2
+ x – 1.
Divisionprocessis shown ontheright side.
We stop here since degree of the remainder is
less than the degree of (–x2
+ x – 1), the divisor.
So, quotient = x – 2, remainder = 3.
Now,
Dividend = Divisor × Quotient + Remainder
= (–x2 + x – 1) (x – 2) + 3
= – x3 + x2 – x + 2x2 – 2x + 2 + 3
= – x3 + 3x2 – 3x + 5
Inthis way, the divisionalgorithmis verified.
2 3 2
3 2
2
2
2
1 3 3 5
2 2 5
2 2 2
3
x
x x x x x
x x x
x x
x x
,
, ∗ , , ∗ , ∗
, ∗ ,
∗ , ∗
, ∗
, ∗
, ∗ ,

Class-X Mathematics
74
Example-11. Find allthe zeroes of2x4
– 3x3
– 3x2
+ 6x – 2, if you know that two of its zeroes
are 2 and – 2 .
Solution : Since two ofthe zeroes are 2 and – 2 , therefore we can divide by
(x – 2 ) (x + 2 ) = x2
– 2.
First termofquotient is
4
2
2
2
2
x
x
x
<
Second termofquotient is
3
2
3
3
x
x
x
,
<,
Third termofquotient is
2
2
1
x
x
<
So, 2x4
– 3x3
– 3x2
+ 6x – 2 = (x2
– 2) (2x2
– 3x + 1).
Now, bysplitting –3x, we factorize 2x2
– 3x + 1 as (2x – 1) (x – 1). So, its zeroes are
given by x =
1
2
and x = 1. Therefore, the zeroes of the given polynomial are 2 , – 2 ,
1 and
1
2
.
EXERCISE – 3.4
1. Dividethepolynomialp(x)bythepolynomialg(x) andfind the quotient andremainder in
eachofthe following :
(i) p(x) = x3
– 3x2
+ 5x – 3, g(x) = x2
– 2
(ii) p(x) = x4
– 3x2
+ 4x + 5, g(x) = x2
+ 1 – x
(iii) p(x) = x4
– 5x + 6, g(x) = 2 – x2
2
2 4 3 2
4 2
3 2
3
2
2
2 3 1
2 2 3 3 6 2
2 4
3 6 2
3 6
2
2
0
x x
x x x x x
x x
x x x
x x
x
x
, ∗
, , , ∗ ,
,
, ∗
, ∗ ∗ ,
, ∗
∗ ,
,
,
, ∗

Polynomials 75
2. Check inwhichcase the first polynomialis a factorofthe second polynomialbydividing
the second polynomialbythe first polynomial:
(i) t2
– 3, 2t4
+ 3t3
– 2t2
– 9t – 12
(ii) x2
+ 3x + 1, 3x4
+ 5x3
– 7x2
+ 2x + 2
(iii) x3
– 3x + 1, x5
– 4x3
+ x2
+ 3x + 1
3. Obtain all other zeroes of 3x4
+ 6x3
– 2x2
– 10x – 5, if two of its zeroes are
5
3
and
5
3
, .
4. On dividing x3
– 3x2
+ x + 2 by a polynomial g(x), the quotient and remainder were
x – 2 and – 2x + 4, respectively. Find g(x).
5. Giveexamplesofpolynomials p(x), g(x),q(x)andr(x),whichsatisfythedivisionalgorithm
and
(i) deg p(x) = deg q(x) (ii) deg q(x) = deg r(x) (iii) deg r(x) = 0
OPTIONAL EXERCISE
1. Verify that the numbers given alongside the cubic polynomials below are their zeroes.
Also verifythe relationship betweenthe zeroes and the coefficientsineach case:
(i) 2x3
+ x2
– 5x + 2 ; (
1
2
, 1, –2) (ii) x3
+ 4x2
+ 5x – 2 ; (1, 1, 1)
2. Find a cubic polynomialwith the sum, sumof the product of its zeroes takentwo at a
time, and the product of its zeroes as 2, –7, –14 respectively.
3. If the zeroes of the polynomial x3
– 3x2
+ x + 1 are a – b, a, a + b find a and b.
4. If two zeroes ofthe polynomial x4
– 6x3
– 26x2
+ 138x – 35 are 2 ± 3 , find the other
zeroes.
5. If the polynomial x4
– 6x3
– 16x2
+ 25x + 10 is divided by another polynomial
x2
– 2x + k, the remainder comes out to be x + a, find k and a.

Class-X Mathematics
76
1. Polynomials of degrees 1, 2 and 3 are called linear, quadratic and cubic polynomials
respectively.
2. Aquadratic polynomialin x withrealcoefficientsis ofthe formax2
+ bx + c, where a,b,
c are realnumbers with a ¹ 0.
3. The zeroes ofa polynomialp(x) arethe x-coordinates ofthe points where the graph of
y = p(x) intersects the X-axis.
4. A quadratic polynomialcan have at most 2 zeroes and a cubic polynomialcan have at
most 3 zeroes.
5. If a and b are the zeroes of the quadratic polynomial ax2
+ bx + c, a ¹ 0, then
a + b = –
b
a
, ab =
c
a
.
6. If a, b, g are the zeroes of the cubic polynomial ax3
+ bx2
+ cx + d, a ¹ 0, then
a + b + g =
b
a
,
,
ab + bg + ga =
c
a
,
and abg =
d
a
,
.
7. The divisionalgorithmstatesthat givenanypolynomialp(x) andanynon-zeropolynomial
g(x), there exist polynomials q(x) and r(x) such that
p(x) = g(x) q(x) + r(x),
where either r(x) = 0 or degree r(x) < degree g(x) if r(x) ¹ 0.
Suggested Projects
Quadratic polynomial - Zeroes of the polynomial - geometrical meaning/ graphs.
l Draw graphs for quadratic polynomialax2
+ bx + c for various conditions.
(i) a > 0 (ii) a < 0 (iii) a = 0 (iv) b > 0 (v) b < 0
(vi) b = 0 and comment onthe graphs

4.1 INTRODUCTION
One day, Siriwent to a book shop with her father andbought 3 notebooks and2 pens. Her
father paid `80 for them. Her friend Laxmiliked the notebooks and pens. So she too bought 4
notebooks and 3 pens of the same kind for `110. Now her classmates Rubina liked the pens
whereas Joseph liked the notebooks. They asked Siri the cost of the pen and the notebook
separately. But, Siri did not know the costs separately. How will they find the costs of these
items?
In this example, the cost of a notebook and a pen are not known. These are unknown
quantities. We comeacross manysuch situations inour day-to-daylife.
THINK - DISCUSS
Two situationsaregivenbelow:
(i) The cost of1kg potatoes and 2kg tomatoes was `30 on a certain day.After two days,
the cost of2kg potatoes and 4kg tomatoes was found to be `66.
(ii) The coach of a cricket team of M.K.Nagar High School buys 3 bats and 6 balls for
`3900. Later he buys one more bat and 2 balls for `1300.
Identifythe unknownsin each situation. We observe that there are two unknowns in each
case.
4.1.1 HOW DO WE FIND UNKNOWN QUANTITIES?
Inthe introduction, Siribought 3 notebooks and 2 pens for `80. Howcan we findthe cost
of anotebook or the cost of a pen?
Rubina and Joseph tried to guess. Rubina said that price ofeach notebook could be `25.
Then three notebooks would cost `75 and the two pens would cost `5. Inthat cases each pen
could be for `2.50.
Joseph felt that `2.50 for one pen was too less. In his opinion, it should be at least `16.
Then the price of each notebook would also be `16.
Pair of Linear Equations
in Two Variables
4

Class-X Mathematics
7 8
We cansee that there can be manypossible valuesfor the price ofa notebook and ofa pen
so that thetotalcost is `80. So, how do we find theprice at which SiriandLaxmibought them?
Byusing onlySiri's situation, wecannot find the costs.We have to use Laxmi's situation also.
4.1.2 USING BOTH EQUATIONS TOGETHER
Laxmi also bought the same types of notebooks and pens as Siri. She paid `110 for 4
notebooks and 3 pens.
So, we have two situations which can berepresented as follows:
(i) Cost of 3 notebooks + 2 pens = `80.
(ii) Cost of 4 notebooks + 3 pens = `110.
Does this help us find the cost of a penand a notebook?
Consider the prices mentioned by Rubina. If the price of one notebook is `25 and the
price of one pen is `2.50 then,
The cost of4 notebooks would be : 4 × 25 = `100
And the cost for 3 pens would be : 3 × 2.50 = `7.50
If Rubina is right then Laxmi should have paid `100 + `7.50 = ` 107.50 but she paid `110.
Now, consider theprices mentioned byJoseph.
The cost of 4 notebooks, if one is for `16, would be : 4 × 16 = ` 64
And the cost for 3pens, ifone is for `16, would be : 3 × 16 = ` 48
IfJoseph is right thenLaxmishould have paid `64 + `48 = `112 but this is more than the
price she paid.
So what do we do? How to find the exact cost of the notebook and the pen?
Ifwe have onlyone equation but two unknowns (variables), we canfind manysolutions.
So, whenwe have two variables, we need at least two independent equations to get a unique
solution. One wayto find the valuesofunknownquantities isbyusing the Modelmethod. Inthis
method, rectangles or portions of rectangles are often used to represent the unknowns. Let us
look at the first situationusingthe modelmethod:
Step-1 : Represent a notebook by and a pen by .
Siribought 3 books and 2 pens for `80.
Laxmibought 4 books and 3 pens for `110.
D80
D110

Pair of Linear Equations in Two Variables 7 9
Step-2:Increase(ordecrease) thequantitiesinproportiontomakeoneofthequantitiesequalin
both situations. Here, we make the number ofpens equal.
In Step 2, we observe a simple proportionalreasoning.
Since Siri bought 3 books and 2 pens for `80, so for 9 books and 6 pens:
3 × 3 = 9 books and 3 × 2 = 6 pens, the cost will be 3 × 80 = `240 (1)
Similarly, Laxmibought 4 books and 3 pens for `110, so:
2 × 4 = 8 books and 2 × 3 = 6 pens will cost 2 × 110 = `220 (2)
After comparing (1) and (2), we can easilyobserve that 1 extra book costs
`240 - ` 220 = `20. So one book cost is `20.
Siri bought 3 books and 2 pens for `80. Since each book costs `20, 3 books cost ` 60.
So the cost of 2 pens become ` 80 - ` 60 = ` 20.
So, cost of each pen is `20 ÷ 2 = `10.
Let us trythese costs in Laxmi's situation. 4 books willcost ` 80 and three pens willcost
` 30 for a total of ` 110, which is true.
Fromtheabovediscussionandcalculation,itisclear thattogetexactlyonesolution(unique
solution) we need at least two independent linear equations in thesame two variables.
In general, an equation of the form ax + by + c = 0, where a, b, c are real numbers and
where at least one of a or b is not zero i.e. 2 2
0
a b
+ ¹ , is called a linear equation in two
variables x and y.
TRY THIS
Markthecorrect optioninthefollowing questions:
1. Whichofthefollowing equations is not alinear equation?
a) 5 + 4x = y+ 3 b) x+2y = y - x
c) 3 - x = y2
+ 4 d) x + y = 0
D240
(3×D80)
(3 books × 3) 9 books (2 pens × 3) 6 pens
D220
(2×D110)
(4 books × 2) 8 books (3 pens × 2) 6 pens

Class-X Mathematics
8 0
2. Whichofthefollowing is a linear equationinone variable?
a) 2x + 1 = y - 3 b) 2t - 1 = 2t + 5
c) 2x - 1 = x2
d) x2
- x + 1 = 0
3. Whichofthe following numbers is a solution for the equation 2(x+ 3)= 18?
a) 5 b) 6 c) 13 d) 21
4. The value ofx whichsatisfies the equation 2x- (4- x) = 5- xis
a) 4.5 b) 3 c) 2.25 d) 0.5
5. The equation x - 4y= 5 has
a) no solution b) uniquesolution
c) two solutions d)infinitelymanysolutions
4.2 SOLUTIONS OF PAIRS OF LINEAR EQUATIONS IN TWO VARIABLES
In the introductoryexample ofnotebooks and pens, how many equations did we have?
We had two equations or a pair oflinear equations in two variables. What do we mean bythe
solution for a pair oflinear equations?
A pair ofvalues ofthe variables xand ywhichtogether satisfyeach one oftheequations
is called asolutionfor a pairoflinear equations.
4.2.1 GRAPHICAL METHOD OF FINDING SOLUTION OF A PAIR OF LINEAR EQUATIONS
What willbethe number ofsolutions for a pair oflinear equations in two variables?Isthe
number ofsolutions infinite or uniqueor none?
In anearlier section, we used the model method for solving the pair oflinear equations.
Now we willuse graphs to solve the equations.
Let: a1
x + b1
y + c1
= 0, (a1
2
+ b1
2
¹ 0) and a2
x + b2
y + c2
= 0; (a2
2
+ b2
2
¹ 0) form a pair
of linear equationin two variables.
The graph of a linear equation in two variables is a straight line. Ordered pairs of real
numbers (x, y) representing pointsonthe line are solutions ofthe equation and ordered pairs of
realnumbers (x, y) that do not represent pointson the line arenot solutions.
Ifwe have two lines inthe same plane, what can be the possible relations between them?
What isthesignificance ofthisrelation?

When two lines are drawn inthe same plane, onlyone of the following three situations is
possible:
i) The two lines mayintersect at one point.
ii) The two lines maynot intersect i.e., theyare parallel.
iii) The two lines maybe coincident.
(actuallyboth are same)
Let us write the equations in the first example in terms of x and y, where x is the cost ofa
notebook and y is the cost of a pen. Then, the equations are 3x + 2y = 80 and 4x + 3y = 110.
For the equation 3x + 2y = 80 For the equation 4x + 3y = 110
x y =
80 3
2
x
-
(x, y) x y =
110 4
3
x
-
(x, y)
0 y =
80 3(0)
2
-
= 40 (0, 40) -10 y =
110 4( 10)
3
- -
= 50 (-10, 50)
10 y =
80 3(10)
2
-
= 25 (10, 25) 20 y =
110 4(20)
3
-
= 10 (20, 10)
20 y =
80 3(20)
2
-
= 10 (20, 10) 50 y =
110 4(50)
3
-
= -30 (50, -30)
30 y =
80 3(30)
2
-
= -5 (30, -5)
After plotting theabove points in the
Cartesian plane, we observe that the two
straight lines are intersecting at the point
(20, 10).
Substituting the values of x and y in
the equations we get 3(20) + 2(10) = 80
and4(20)+3(10)=110.Showingtheboth
theequations satisfying.
Thus, as determined bythe graphical
method, the cost ofeach book is `20 and
of each pen is `10. Recallthat we got the
same solutionusingthe modelmethod.
Since (20, 10) is the common point,
thereisonlyonesolutionforthispairoflinear
equations in two variables. Such equations
are known as consistent and independent
pairs oflinear equations. Theywillalways
have aunique solution.
X
5 10
-5
-10 0 15 20 25 30 35 40 45 50
5
10
15
20
25
30
35
40
45
50
55
-5
-10
-15
-20
-25
-30
55
60
Y
X
I
Y
|
(20, 10)
4
x
+
3
y
=
1
1
0
3
x
+
2
y
=
8
0

Class-X Mathematics
8 2
Now, let uslook at the first example fromthe ‘think and discuss section’.Wewant to find
the cost of1kg ofpotatoes andthe cost of1kg oftomatoes each. Let the cost of1kg potatoes
be `x and cost of 1kg of tomato be `y. Then, the equations will become 1x+2y=30 and
2x+4y=66.
For the equation x + 2y = 30 For the equation 2x + 4y = 66
x y =
30
2
x
-
(x, y) x y =
66 2
4
x
-
(x, y)
0 y =
30 0
2
-
= 15 (0, 15) 1 y =
66 2(1)
4
-
= 16 (1, 16)
2 y =
30 2
2
-
= 14 (2, 14) 3 y =
66 2(3)
4
-
= 15 (3, 15)
4 y =
30 4
2
-
= 13 (4, 13) 5 y =
66 2(5)
4
-
= 14 (5, 14)
6 y =
30 6
2
-
= 12 (6, 12) 7 y =
66 2(7)
4
-
= 13 (7, 13)
Here, we observe that the situation is
represented graphicallybytwo parallel lines. Since
the lines do not intersect, the equations have no
common solution. This means that the cost of the
potato and tomato was different on different days.
We see this in real life also. We cannot expect the
samepriceofvegetableseveryday;itkeepschanging.
Also, the change is independent.
Such pairs of linear equations which have no
solution are known as inconsistent pairs of linear
equations.
In the second example from the think and
discuss section, let the cost of each bat be ` x and
each ballbe `y. Thenwe can write theequations as
3x + 6y = 3900 and x + 2y = 1300.
For the equation 3x + 6y = 3900 For the equation x + 2y = 1300
x y =
3900 3
6
x
-
(x, y) x y =
1300
2
x
-
(x, y)
100 y =
3900 3(100)
6
-
= 600 (100, 600) 100 y =
1300 100
2
-
= 600 (100, 600)
2x + 4y = 66
x + 2y = 30
X
1
-1 0 2 3 4 6
5 7 8
6
7
8
9
10
11
12
13
14
15
-1
-2
16
17
Y
X
I
Y
|
-2
1
2
3
4
5
Scale
xaxis: 1 cm=1 unit
yaxis: 1cm= 1 unit

200 y =
3900 3(200)
6
-
= 550 (200, 550) 200 y =
1300 200
2
-
= 550 (200, 550)
300 y =
3900 3(300)
6
-
= 500 (300, 500) 300 y =
1300 300
2
-
= 500 (300, 500)
400 y =
3900 3(400)
6
-
= 450 (400, 450) 400 y =
1300 400
2
-
= 450 (400, 450)
The equations are geometrically shown
bya pair ofcoincident lines. Ifthe solutions of
the equations are given bythe commonpoints,
then what are the common points in this case?
From the graph, we observe that every
point on the line is a commonsolution to both
the equations. So, they have infinitely many
solutions as boththe equations are equivalent.
Such pairs of equations are called consistent
and dependent pair oflinearequationsintwo
variables. This system of equations that has
solutionare knownas‘consistent equations’.
TRY THIS
In the example given above, can you find the cost ofeachbat and ball?
THINK - DISCUSS
Is adependent pair oflinear equations always consistent. Whyor whynot?
DO THIS
1. Solve the followingsystems ofequations :
i) x - 2y = 0 ii) x + y = 2 iii) 2x - y = 4
3x + 4y = 20 2x + 2y = 4 4x - 2y = 6
2. Represent the following pair oflinear equationsgraphically.
x + 2y - 4 = 0 and 2x + 4y- 12 = 0. Represent this situation graphically.
4.2.3 RELATION BETWEEN COEFFICIENTS AND NATURE OF SYSTEM OF EQUATIONS
Let a1
, b1
, c1
and a2
, b2
, c2
denote thecoefficients of a given pair oflinear equations intwo
variables. Then, let us write and compare the valuesof
1 1
2 2
,
a b
a b and
1
2
c
c inthe above examples.
X
100
-100 0 200 300 400 600
500
600
-100
-200
Y
X
I
Y
|
-200
100
200
300
400
500
700
Scale
x axis: 1 cm=100 units
yaxis: 1 cm= 100 units

Class-X Mathematics
8 4
Pair oflines
1
2
a
a
1
2
b
b
1
2
c
c Comparison Algebraic Graphical Solutions
ofratios interpretation representation
1. 3x+2y–80=0
3
4
2
3
80
110
-
-
1 1
2 2
a b
a b
¹ Consistent and Intersecting One
4x+3y–110=0 Independent solution
2. 1x+2y–30=0
1
2
2
4
30
66
-
-
1 1 1
2 2 2
a b c
a b c
= ¹ Inconsistent Parallel No
2x+4y–66=0 solution
3. 3x+6y=3900
3
1
6
2
3900
1300
1 1 1
2 2 2
a b c
a b c
= = Consistent Coincident Infinitely
x+2y=1300 and many
dependent solutions
Let us look at examples.
Example-1. Check whether the given pair of equations represent intersecting, parallel or
coincident lines. Find the solution ifthe equations are consistent.
2x + y - 5 = 0
3x - 2y - 4 = 0
Solution :
1
2
2
3
a
a
= 1
2
1
2
b
b
=
-
1
2
5
4
c
c
-
=
-
Since
1 1
2 2
a b
a b
¹ , thereforetheyare intersecting lines and hence, it is a consistent pair of linear
equation.
For the equation 2x + y = 5 For the equation 3x - 2y = 4
x y = 5 - 2x (x, y) x y =
4 3
2
x
-
-
(x, y)
0 y = 5 - 2 (0) = 5 (0, 5) 0 y =
4 3(0)
2
-
-
= -2 (0, -2)
1 y = 5 - 2(1) = 3 (1, 3) 2 y =
4 3(2)
2
-
-
= 1 (2, 1)
2 y = 5 - 2(2) = 1 (2, 1) 4 y =
4 3(4)
2
-
-
= 4 (4, 4)
3 y = 5 - 2(3) = -1 (3, -1)
4 y = 5 - 2(4) = -3 (4, -3)

X
1
-1 0 2 3 4 6
5 7 8
6
7
8
-1
-2
Y
X
I
Y|
-2
1
2
3
4
5
-3
(2, 1)
Theuniquesolutionof
this pairofequations
is (2,1).
Example-2. Check whetherthe following pair ofequations is consistent.
3x + 4y = 2 and 6x + 8y = 4. Verify by a graphical representation.
Solution : 3x + 4y - 2 = 0
6x+8y - 4 = 0
1
2
3 1
6 2
a
a
= = 1
2
4 1
8 2
b
b
= = 1
2
2 1
4 2
c
c
-
= =
-
Since
1 1 1
2 2 2
a b c
a b c
= = , therefore, theyare coincident lines. So, the pair oflinear equations
are consistent and dependent and haveinfinitelymanysolutions.
For the equation 3x + 4y = 2 For the equation 6x + 8y = 4
x y =
2 3
4
x
-
(x, y) x y =
4 6
8
x
-
(x, y)
0 y =
2 3(0)
4
-
=
1
2
(0,
1
2
) 0 y =
4 6(0)
8
-
=
1
2
(0,
1
2
)
2 y =
2 3(2)
4
-
= -1 (2, -1) 2 y =
4 6(2)
8
-
= -1 (2, -1)
4 y =
2 3(4)
4
-
= -2.5 (4, -2.5) 4 y =
4 6(4)
8
-
= -2.5 (4, -2.5)
6 y =
2 3(6)
4
-
= -4 (6, -4) 6 y =
4 6(6)
8
-
= -4 (6, -4)
2
x
+
y
=
5
3
x
-
2
y
=
4
Scale
xaxis: 1 cm=1 unit
yaxis: 1 cm= 1 unit

Class-X Mathematics
8 6
Scale
xaxis: 1 cm=1 unit
yaxis: 1 cm= 1 unit
Example-3. Check whether the equations 2x-3y= 5and 4x-6y= 15 are consistent.Also verify
bygraphicalrepresentation.
Solution : 4x-6y - 15 = 0
2x-3y - 5 = 0
1
2
4 2
2 1
a
a
= = 1
2
6 2
3 1
b
b
-
= =
-
1
2
15 3
5 1
c
c
-
= =
-
1 1 1
2 2 2
a b c
a b c
= ¹
So theequationsare inconsistent.Theyhaveno solutionsandtheir graphisofparallellines.
For the equation 4x - 6y = 15 For the equation 2x - 3y = 5
x y =
15 4
6
x
-
-
(x, y) x y =
5 2
3
x
-
-
(x, y)
0 y =
15 0 5
6 2
- -
=
-
(0, -2.5) 1 y =
5 2(1)
1
3
-
= -
-
(1, -1)
3 y =
15 4(3) 1
6 2
- -
=
-
(3, -0.5) 4 y =
5 2(4)
1
3
-
=
-
(4, 1)
6 y =
15 4(6) 3
6 2
-
=
-
(6, 1.5) 7 y =
5 2(7)
3
3
-
=
-
(7, 3)
X
1
-1 0 2 3 4 6
5 7 8
-1
-2
Y
XI
Y
|
-2
1
2
3
4
-3
-4
Scale
xaxis: 1 cm=1 unit
yaxis: 1 cm= 1 unit

DO THIS
Check each of the given systems of equations to see if it has a unique solution,
infinitelymanysolutionsor no solution.Solve themgraphically.
(i) 2x+3y = 1 (ii) x + 2y = 6 (iii) 3x + 2y = 6
3x-y = 7 2x + 4y = 12 6x + 4y = 18
TRY THIS
1. For what valueof'p'the followingpair ofequations hasa unique solution.
2x + py = - 5 and 3x + 3y = - 6
2. Find the value of 'k' for which the pair of equations 2x - ky + 3 = 0, 4x + 6y - 5 =0
represent parallellines.
3. For what value of 'k', the pair of equations 3x + 4y + 2 = 0 and 9x + 12y + k = 0
represents coincident lines.
4. For what positivevaluesof'p', thefollowing pair oflinear equationshave infinitelymany
solutions?
px + 3y - (p - 3) = 0
12x + py - p = 0
Let us look at some more examples.
Example-4. In a garden there are some bees and flowers. Ifone bee sits oneach flower then
one bee willbe left. If two bees sit oneach flower, one flower will be left. Find the number of
bees and number offlowers.
Solution : Let the number of bees = x and
the number offlowers = y
If one bee sits on each flower then one bee willbe left. So, x = y + 1
X
1
-1 0 2 3 4 6
5 7 8
-1
-2
Y
X
I
Y
|
-2
1
2
3
4
-3
-4
5
4x - 6y = 9
2x - 3y = 5
Scale
xaxis: 1 cm=1 unit
yaxis: 1 cm= 1 unit

Class-X Mathematics
8 8
or x - y - 1 = 0 ... (1)
If two bees sit on each flower, one flower willbe left. So, x = 2(y- 1)
or x - 2y+2 = 0 ... (2)
For the equation x - y - 1 = 0 For the equation x - 2y + 2 = 0
x y = x - 1 (x, y) x y =
2
2
x +
(x, y)
0 y = 0 - 1 = -1 (0, -1) 0 y =
0 2
2
+
= 1 (0, 1)
1 y = 1 - 1 = 0 (1, 0) 2 y =
2 2
2
+
= 2 (2, 2)
2 y = 2 - 1 = 1 (2, 1) 4 y =
4 2
2
+
= 3 (4, 3)
3 y = 3 - 1 = 2 (3, 2) 6 y =
6 2
2
+
= 4 (6, 4)
4 y = 4 - 1 = 3 (4, 3)
Scale
xaxis: 1 cm=1 unit
yaxis: 1 cm= 1 unit
In the graph, (4, 3) is the point ofintersection. Therefore, there are 4 bees and 3 flowers.
Example-5. The perimeter ofa rectangular plot is32m. Ifthelengthis increased by2mand the
breadth is decreased by1m, the areaofthe plot remains the same. Find thelength and breadth
ofthe plot.
Solution : Let length and breadthofthe rectangular land be l and brespectively. Then,
area = lb and
Perimeter = 2(l + b) = 32 m.
Then, l + b = 16 which implies l + b - 16 = 0 ... (1)
X
1
-1 0 2 3 4 6
5 7 8
-1
-2
Y
XI
Y|
-2
1
2
3
4
-3
-4
5
(4, 3)
x
-
y
-
1
=
0
x - 2y + 2 = 0
Scale
xaxis: 1 cm=1 unit
yaxis: 1cm= 1 unit

When length is increased by 2m., then new length is l + 2.Also breadth is decreased by
1m; so new breadth is b - 1.
Then, area = (l + 2) (b - 1)
Since thereis no change inthe area,
(l + 2) (b - 1) = lb
lb - l +2b - 2 = lb or lb - lb = l - 2b + 2
l - 2b + 2 = 0 ... (2)
For the equation l + b - 16 = 0 For the equation l - 2b + 2 = 0
l b = 16 - l (l, b) l b =
2
2
l +
(l, b)
6 b = 16 - 6 = 10 (6, 10) 6 b =
6 2
2
+
= 4 (6, 4)
8 b = 16 - 8 = 8 (8, 8) 8 b =
8 2
2
+
= 5 (8, 5)
10 b = 16 - 10 = 6 (10, 6) 10 b =
10 2
2
+
= 6 (10, 6)
12 b = 16 - 12 = 4 (12, 4) 12 b =
12 2
2
+
= 7 (12, 7)
14 b = 16 - 14 = 2 (14, 2) 14 b =
14 2
2
+
= 8 (14, 8)
So, originallength ofthe plot is 10mand its breadth is 6m.
Taking measures of length on X-axis and measure of breadthonY-axis, we get the graph
X
2
-1 0 4 6 8 12
10 14 16
-1
-2
Y
XI
Y|
-2
2
4
6
8
-3
-4
10
12
14
16 18
(10, 6)
l - 2b + 2 = 0
l +
b
- 16
=
0
Scale
xaxis: 1 cm=1 unit
yaxis: 1 cm= 1 unit

Class-X Mathematics
9 0
EXERCISE - 4.1
1. Bycomparingtheratios
1 1 1
2 2 2
, ,
a b c
a b c ,statewhetherthelinesrepresentedbythefollowing
pairs oflinear equations intersect at a point, are parallelor are coincident.
a) 5x- 4y + 8 = 0 b) 9x+3y + 12 = 0 c) 6x - 3y + 10 = 0
7x+6y-9 = 0 18x+6y + 24 = 0 2x - y + 9 = 0
2. Checkwhetherthefollowingequationsareconsistent orinconsistent.Solvethemgraphically.
a) 3x+2y=5 b) 2x - 3y = 8 c)
3 5
2 3
x y
+ = 7
2x - 3y=7 4x - 6y = 9 9x - 10y = 12
d) 5x-3y = 11 e)
4
3
x +2y = 8 f) x + y = 5
-10x+6y = -22 2x+3y = 12 2x+2y = 10
g) x - y = 8 h) 2x + y-6 = 0 i) 2x-2y - 2 = 0
3x-3y = 16 4x-2y- 4 = 0 4x-4y- 5 = 0
3. Neha went to a 'sale' to purchase some pants and skirts. When her friend asked her how
manyofeachshe had bought, she answered, "the number ofskirts are two less thantwice
the number of pants purchased and the number of skirts is four less thanfour times the
number ofpants purchased."
Help her friend to find howmanypants and skirts Neha bought.
4. 10 students ofClass-X took part ina mathematics quiz. Ifthe number of girls is 4 more
than the number ofboys then, find the number ofboys and the number ofgirls who took
part inthe quiz.
5. 5 pencils and 7 pens together cost `50 whereas 7 pencils and 5 pens together cost D46.
Find the cost ofone pencil and that ofone pen.
6. Half the perimeter ofarectangular gardenis 36m. If thelengthis 4mmore than its width,
find the dimensionsofthe garden.
7. We have a linear equation 2x + 3y - 8 = 0. Write another linear equation intwo variables
x and y such that the geometrical representation ofthe pair so formedis intersecting lines.
Now, write two more linear equations so that one forms a pair of parallel lines and the
second formscoincident line withthegiven equation.
8. The areaofa rectangle gets reduced by80 sq units ifits lengthis reduced by5 units and
breadth is increased by 2 units. If we increase the length by 10 units and decrease the

breadthby5 units, the area willincrease by50 sq units. Find the length and breadthofthe
rectangle.
9. In a class, ifthreestudentssit oneachbench,one student willbe left. Iffourstudentssit on
eachbench,one benchwillbeleft. Findthe numberofstudentsand thenumber ofbenches
inthat class.
4.3 ALGEBRAIC METHODS OF FINDING THE SOLUTIONS FOR A PAIR OF LINEAR
EQUATIONS
Wehavelearnt howto solveapair oflinearequations graphically.But,thegraphicalmethod
is not convenient in cases where thepoint representing the solution hasno integralco-ordinates.
For example, when the solutionis ofthe form( 3 , 2 7 ), (- 1.75, 3.3), (
4
13
,
1
19
) etc. There
is everypossibilityof making mistakes while reading such co-ordinates. Is there anyalternative
method offinding a solution?There are severalother methods, some ofwhich we shalldiscuss
now.
4.3.1 SUBSTITUTION METHOD
This method is useful for solving a pair of linear equations in two variables where one
variable caneasily be written in terms of the other variable. To understand this method, let us
consider it step-wise.
Step-1 : In one of the equations, express one variable in terms ofthe other variable. Say y in
terms of x.
Step-2 : Substitute the value ofy obtained in step 1 inthe second equation.
Step-3 : Simplifythe equation obtained in step 2 and find the value ofx.
Step-4 : Substitute the value of x obtainedinstep 3 ineither oftheequations and solve it for y.
Step-5 : Check the obtained solution bysubstituting the values of x and y inboth the original
equations.
Example-6. Solve thegivenpair ofequationsusing substitutionmethod.
2x - y = 5
3x + 2y = 11
Solution : 2x - y = 5 (1)
3x + 2y = 11 (2)
Equation(1) canbe written as
y = 2x - 5
Substituting inequation(2) we get

Class-X Mathematics
9 2
3x + 2(2x - 5) = 11
3x + 4x - 10 = 11
7x = 11 + 10 = 21
x = 21/7 = 3.
Substitute x =3 in equation (1)
2(3) - y = 5
y = 6 - 5 = 1
Substituting the values of x and y in equation (2), we get 3(3) + 2(1) = 9 + 2 = 11
Both the equations are satisfied by x = 3 and y = 1.
Therefore, the required solution is x = 3 and y = 1.
DO THIS
Solve following pair ofequationsbyusingthe substitutionmethod.
1) 3x - 5y = -1 2) x+2y = - 1 3) 2x+3y = 9
x - y = - 1 2x - 3y = 12 3x+4y = 5
4)
6
x
y
+ = 6 5) 0.2x + 0.3y = 13 6) 2 + 3 = 0
x y
8
3x
y
- = 5 0.4x + 0.5y = 2.3 3 - 8 = 0
x y
4.3.2 ELIMINATION METHOD
Inthismethod,firstweeliminate(remove)oneofthetwo variablesbyequatingitscoefficients.
Thisgivesasingleequationwhichcanbesolvedto getthevalueoftheothervariable.Tounderstand
this method, let us consider it stepwise.
Step-1 : Write both the equations in the form of ax + by = c.
Step-2 : Makethecoefficientsofoneofthevariables,say'x', equalbymultiplying eachequation
bysuitablerealnumbers.
Step-3 : Ifthevariableto beeliminatedhasthesamesigninbothequations,subtractoneequation
fromthe other to get anequationinone variable. Iftheyhave opposite signs then add.
Step-4 : Solve the equationfor the remaining variable.
Step-5 : Substitute the value ofthis variable in any one ofthe originalequations and find the
value ofthe eliminated variable.
Example-7. Solvethe followingpair oflinear equationsusing eliminationmethod.
3x + 2y = 11
2x + 3y = 4

Solution : 3x + 2y = 11 (1)
2x + 3y = 4 (2)
Let us eliminate 'y' fromthe given equations. The coefficients of'y'in the given equations are 2
and 3. L.C.M.of 2 and 3 is 6. So, multiply equation (1) by3 and equation(2) by2.
Equation (1) × 3 9x + 6y = 33
Equation (2) × 2 4x + 6y = 8
(-) (-) (-)
5x = 25
x =
25
5
= 5
Substitute x = 5, in equation (1)
3(5) + 2y = 11
2y = 11 - 15 = - 4
4
2
2
-
Þ = = -
y
Therefore, the required solution is x = 5, y = - 2.
DO THIS
Solve eachofthe following pairsofequations bytheeliminationmethod.
1. 8x+ 5y = 9 2. 2x + 3y = 8 3. 3x + 4y = 25
3x+2y = 4 4x + 6y = 7 5x - 6y = -9
TRY THIS
Solve the givenpair of linear equations
(a - b)x+ (a + b)y = a2
- 2ab - b2
(a + b) (x + y) = a2
+ b2
Let us see some more examples:
Example-8. Rubina went to abank to withdraw `2000. She asked the cashier to give the cash
in `50 and `100 notes only. She got 25 notes in all. Can you tellhow manynotes each of `50
and `100 she received?
Solution : Let the number of `50 notes be x;
Let the number of `100 notesbe y;
then, x + y = 25 (1)
and 50x + 100y = 2000 (2)
Solutionsthroughthesubstitutionmethod:

Class-X Mathematics
9 4
Fromequation (1) x = 25 -y
Substituting inequation (2) 50 (25 - y) + 100y = 2000
1250 - 50y + 100y = 2000
50y = 2000 - 1250 = 750
y =
750
50
= 15
x = 25 - 15 = 10
Hence, Rubina received ten `50 notes and fifteen `100 notes.
Solutionthroughtheeliminationmethod:
In the equations, coefficients ofx are 1 and 50 respectively. So,
Equation (1) × 50 : 50x + 50y = 1250
Equation (2) × 1 : 50x + 100y = 2000 same sign, so subtract
(-) (-) (-)
-50y = -750
or y =
750
50
-
-
= 15
Substitute yinequation (1) x + 15 = 25
x = 25 - 15 = 10
Hence Rubina received ten D50 notes and fifteen D100 rupee notes.
Example-9. In a competitive exam, 3 marks are awarded for every correct answer and for
every wrong answer, 1 mark willbe deducted. Madhu scored 40 marks in this exam. Had 4
marks been awarded for each correct answer and 2 marks deducted for each incorrect answer,
Madhu would havescored50 marks. IfMadhu hasattempted allquestions, how manyquestions
were therein the test?
Solution : Let the number ofcorrect answers be x
and the number ofwrong answers be y.
When 3 marks are given for each correct answer and 1 mark deducted for each wrong
answer, his score is 40 marks.
So 3x - y = 40 (1)
His score would have been 50 marks if 4 marks were givenfor eachcorrect answer and 2
marks deducted foreach wrong answer.
Thus, 4x - 2y = 50 (2)

Substitution method
Fromequation(1), y = 3x - 40
Substituting inequation (2) 4x - 2 (3x - 40) = 50
4x - 6x + 80 = 50
- 2x = 50 - 80 = -30
x =
30
2
-
-
=15
Substitute thevalue ofxinequation (1)
3(15) - y = 40
45 - y = 40
y = 45 - 40 = 5
Total number of questions = 15 + 5 = 20
DO THIS
Use the eliminationmethod to solvethe example-9.
Example-10. Marytold her daughter, "Sevenyears ago, I was seven times as oldas you were
then. Also, three years fromnow, I shallbe three times as oldas you willbe." Find the present
age ofMaryand her daughter.
Solution : Let Mary's present age be x years and her daughter's age be y years.
Then, seven years ago Mary's age was x - 7 and daughter's age was y - 7.
x - 7 = 7(y - 7)
x - 7 = 7y - 49
x - 7y + 42 = 0 (1)
Three years hence, Mary's age will be x + 3 and daughter's age will be y + 3.
x + 3 = 3 (y + 3)
x + 3 = 3y + 9
x - 3y - 6 = 0 (2)
Elimination method
Equation1 x - 7y = - 42
Equation2 x - 3y = 6
(-) (+) (-) same sign for x, so subtract.
-4y = -48

Class-X Mathematics
9 6
y =
48
4
-
-
= 12
Substitute the value ofy in equation (2)
x-3 (12) - 6 = 0
x = 36 + 6 = 42
Therefore, Mary's present age is 42 years and her daughter's age is 12 years.`
DO THIS
Solve example-10 bythe substitution method.
Example-11. Apublisher is planning to produce a new
textbook. The fixedcosts(reviewing, editing, typesetting
and so on) are ` 320000. Besides that, he also spends
another ` 31.25 in producing the book. The wholesale
price (the amount received by the publisher) is ` 43.75
per book. How many books must the publisher sell to
break even, i.e., so that the cost ofproduction will equal
revenues?
Solution : The publisher breakseven whencosts equalrevenues. If x representsthe number of
books printed andsold and y bethe breakevenpoint, thenthe cost and revenueequationsfor the
publisher are
Cost equation is given by y = 320000 + 31.25x (1)
Revenue equationisgivenby y = 43.75x (2)
Using the secondequationto substitute for yin the first equation, we have
43.75x = ` 320000 + 31.25x
12.5x = ` 320000
x =
320000
12.5
= 25,600
Thus, the publisher will break evenwhen 25,600 books are printed and sold.
EXERCISE - 4.2
Forma pair of linear equationsfor eachofthefollowing problems and findtheir solution.
1. The ratio ofincomes oftwo persons is 9 : 7 and the ratio oftheir expenditures is 4 : 3. If
each of themmanages to save `2000 per month,find their monthlyincome.
The point which corresponds to
howmuchmoneyyouhavetoearn
throughsalesinorder to equalthe
moneyyou spent inproductionis
break even point.

2. The sumofa two digit number and the number obtained byreversing the digits is66. Ifthe
digits ofthe number differ by2, find the number. How manysuch numbers are there?
3. The larger oftwo supplementaryangles exceeds the smaller by18°. Find the angles.
4. The taxicharges in Hyderabad are fixed, along with the charge for the distance covered.
Upto first 3 kmyou willbe charged a certainminimumamount. Fromthere onwardsyou
have to pay additionallyfor everykilometer travelled. For thefirst 10 km, the charge paid
is `166. For a journey of 15 km. the charge paid is `256.
i. What are the fixed charges and charge per km?
ii. How much does a person have to payfor travelling a distance of25 km?
5. Afractionwillbe equalto
4
5
if1 isadded tobothnumerator and denominator.If, however,
,
5 is subtracted fromboth numerator and denominator, the fraction will be equal to
1
2
.
What isthe fraction?
6. PlacesAand B are 100kmapart on ahighway. One car startsfromAand another fromB
at the same time at different speeds. Ifthe carstravelin the same direction, theymeet in 5
hours. If theytravel towards each other, they meet in 1 hour. What are the speeds of the
two cars?
7. Two angles are complementary. The larger angle is 3° lessthan twice the measure of the
smaller angle. Find the measure ofeachangle.
8. Adictionaryhas a totalof1382pages. It isbrokenup into two parts. The secondpartofthe
book has 64pages more thanthefirst part. How manypages are ineachpart of the book?
9. A chemist has two solutions of hydrochloric acid in stock. One is 50% solution and the
other is 80% solution. How much of each should be used to obtain 100ml of a 68%
solution.
10. You have `12,000/- saved amount, and wants to invest it in two schemes yielding 10%
and 15% interest. How much amount should be invested in each scheme so that you
should get overall12% interest.
4.4 EQUATIONS REDUCIBLE TO A PAIR OF LINEAR EQUATIONS IN TWO
VARIABLES
Now we shall discuss the solution of pairs of equations which are not linear but can be
reduced to linear form bymakingsuitable substitutions. Let ussee anexample:
Example-12. Solve the following pairofequations.
2 3
x y
+ = 13
5 4
x y
- = -2
Solution : Observe the givenpair ofequations. They are not linear equations. (Why?)

Class-X Mathematics
9 8
We have 2
1
x
æ ö
ç ÷
è ø
+ 3
1
y
æ ö
ç ÷
è ø
= 13 (1)
5
1
x
æ ö
ç ÷
è ø
- 4
1
y
æ ö
ç ÷
è ø
= -2 (2)
If we substitute
1
x
= p and
1
y
= q, we get the following pair oflinear equations:
2p + 3q = 13 (3)
5p - 4q = -2 (4)
Coefficients of q are 3 and4 and their LCM is 12. Using theeliminationmethod:
Equation (3) × 4 8p + 12q = 52
Equation (4) × 3 15p - 12q = -6 'q' terms haveopposite sign, so we add the two equations.
23p= 46
p =
46
23
= 2
Substitute thevalue ofp inequation (3)
2(2) +3q = 13
3q = 13 - 4 = 9
q =
9
3
= 3
But,
1
x
= p = 2 Þ x =
1
2
1
y
= q = 3 Þ y =
1
3
Example-13. Kavitha thought ofconstructing 2 more rooms in her house. She enquired about
the labour. She came to knowthat 6 menand8 womencould finishthis work in 14 days. But she
wishes to complete that work in only10 days. When she enquired, she was told that 8 men and
12 womencould finishthework in10 days. Find out how muchtime would betakento finishthe
work ifone man or one woman worked alone.
Solution : Let the time taken by one man to finish the work = x days.

The portion ofwork done byone manin one day =
1
x
Let the time takenbyone womanto finishthe work = y days.
The portion ofwork done byone womanin one day =
1
y
Now, 8 men and 12 womencan finish the work in 10 days.
So the portion of work done by8 men and 12 women in one day =
1
10
(1)
Also, the portion of work done by8 men in one day is 8 ×
1
x
. =
8
x
Similarly, the portion ofwork done by12 women in one dayis 12 ×
1
y
=
12
y
Totalportion ofwork done by8 men and 12 womenin one day =
8 12
x y
+ (2)
Equating equations (1) and (2)
8 12 1
10
x y
æ ö
+ =
ç ÷
è ø
10
8 12
x y
æ ö
+
ç ÷
è ø
= 1
80 120
x y
+ = 1 (3)
Also, 6 men and 8 womencan finish the work in 14 days.
The portionof work done by6 men and 8 women in one day=
6 8 1
14
x y
+ =
Þ 14
6 8
x y
æ ö
+
ç ÷
è ø
= 1
84 112
x y
æ ö
+
ç ÷
è ø
= 1 (4)

Class-X Mathematics
100
Observe equations (3) and (4).Are theylinear equations? How do we solve themthen?We can
convert them into linear equations by substituting
1
x
= u and
1
y
= v.
Equation (3) becomes 80u + 120v = 1 (5)
Equation (4) becomes 84u + 112v = 1 (6)
L.C.M. of80 and 84 is 1680. Using the elimination method,
Equation (3) × 21 (21 × 80)u + (21 × 120)v = 21
Equation (4) × 20 (20 × 84)u + (20 × 112)v = 20
1680u+2520v = 21
1680u+2240v = 20 Same sign for u, so subtract
(-) (-) (-)
280v = 1
v =
1
280
Substitute inequation(5) 80u + 120 ×
1
280
= 1
80u = 1 -
3
7
=
7 3
7
-
=
4
7
1
4 1
=
7 80
´
u
20
1
=
140
So one man alone can finish the work in 140 days and one woman alone can finish the
work in 280 days.
Example-14. Aman travels 370 km partly by train and partly by car. If he covers 250 km by
train andthe rest bycar, it takes him 4hours. But ifhetravels 130 km bytrain and the rest bycar,
it takes18 minutes more. Find the speed of the train and that ofthe car.
Solution : Let the speed of the train be x km. per hour and that of the car be y km. per hour.
Also, we know that time =
Distance
Speed
Insituation1, time spent travelling bytrain =
250
x
hrs.
And time spent travelling bycar =
120
y
hrs.

Pair of Linear Equations in Two Variables 101
So, totaltime taken = time spent in train + time spent in car =
250 120
+
x y
But, totaltime ofjourneyis 4 hours, so
250 120
+ = 4
x y
125 60
+ = 2
x y
® (1)
Again, whenhe travels 130 kmby train and the rest by car
Time takenbyhimto travel130 km bytrain =
130
x
hrs.
Time taken by him to travel 240 km (370 - 130) by car =
240
y
hrs.
Totaltime taken =
130 240
+
x y
But given, time ofjourneyis 4 hrs 18 min i.e.,
18
4
3
60
10
hrs. =
3
4
10
hrs.
So,
130 240 43
+ =
10
x y
(2)
Substitute
1
x
= a and
1
y
= b in equations (1) and (2)
125a + 60b = 2 (3)
130a+ 240b = 43/10 (4)
For 60 and 240, LCM is 240. Using the elimination method,
Equation (3) × 4 500a+240b= 8
Equation (4) × 1 130a+240b =
43
10
(Same sign, so subtract)
(-) (-) (-)
370a = 8 -
43 80 43 37
= =
10 10 10
-

Class-X Mathematics
102
a =
37 1
10 370
´
10
1
=
100
Substitute a =
1
100
inequation (3)
125
5
1
100
´
4
æ ö
ç ÷
ç ÷
è ø
+ 60b = 2
60b = 2 -
5 8 5 3
= =
4 4 4
-
b =
3 1
4 60
´
20
1
=
80
So a =
1
100
and b =
1
80
So
1 1
=
100
x
and
1 1
=
80
y
x = 100 km/hr and y = 80 km/hr.
So, speed of the train was 100 km/hr and speedof the car was 80 km/hr.
EXERCISE - 4.3
Solveeachofthefollowingpairsofequationsbyreducingthemto apairoflinearequations.
i)
5 1
+
1 2
x y
- -
= 2 ii) = 2
+
x y
xy
6 3
-
1 2
x y
- -
= 1
-
x y
xy
= 6
iii)
2 3
+
x y = 2 iv) 6x+3y= 6xy
4 9
-
x y = -1 2x + 4y = 5xy

Pair of Linear Equations in Two Variables 103
v)
5 2
-
x y x y
+ -
= -1 vi)
2 3
+
x y
= 13
15 7
+
x y x y
+ -
= 10
5 4
-
x y
= -2
vii)
10 2
+
x y x y
+ -
= 4 viii)
1 1 3
+ =
3 3 4
x y x y
+ -
15 5
-
x y x y
+ -
= -2
( )
1 1 1
- =
2 3 2(3 ) 8
x y x y
-
+ -
2. Formulate the following problems as a pair of equations andthenfind their solutions.
i. A boat goes 30 km upstreamand 44 kmdownstream in10 hours. In 13 hours it can
go 40 kmupstream and 55 kmdownstream. Determine the speed ofthe stream and
that ofthe boat in stillwater.
ii. Rahim travels 600 kmto his homepartly bytrain and partlybycar. Hetakes 8 hours
if he travels 120 km by train and rest bycar. He takes 20 minutes more if he travels
200 kmbytrain and rest by car. Find the speed ofthe train and the car.
iii. 2 womenand 5 men cantogether finishan embroiderywork in 4 dayswhile 3 women
and 6 men can finish it in 3 days. Find the time to be taken by 1 woman alone and 1
man alone to finishthe work.
OPTIONAL EXERCISE
1. Solve thefollowing equations:-
(i)
2
+
x y
a b
= 2 (ii)
1 1
+
2 3
x y
+ -
= 8
-
x y
a b
= 4
1 1
+
3 2
x y
- +
= 9
(iii) +
7 3
x y
= 5 (iv) 3 + 2 = 3
x y
-
2 9
x y
= 6 5 + 3 = 3
x y
(v) -
ax by
b a
= a + b (vi) 2x
+ 3y
= 17
ax - by = 2ab 2x+2
- 3y+1
= 5

Class-X Mathematics
104
2. Animals in anexperiment are to be kept on a strict diet. Each animalis to receive among
other things 20gofprotein and 6goffat. The laboratorytechnicianspurchased two food
mixes,Aand B. MixAhas 10% protein and 6% fat. Mix B has 20% protein and 2% fat.
How manygrams ofeach mixshould be used?
1. Two linear equations in the same two variables are called a pair of linear equations intwo
variables.
a1
x + b1
y + c1
= 0 (a1
2
+ b1
2
¹ 0)
a2
x + b2
y + c2
= 0 (a2
2
+ b2
2
¹ 0)
where a1
, a2
, b1
, b2
, c1
, c2
are real numbers.
2. A pair oflinear equations intwo variables canbe solvedusing various methods.
3. The graphofa pair oflinear equations in two variablesis represented bytwo lines.
i. Ifthelinesintersectatapointthenthepointgivestheuniquesolutionofthetwoequations.
In this case, the pair ofequations isconsistent.
ii. If the lines coincide, then there are infinitely many solutions - each point on the line
being a solution. In this case, the pair ofequations is dependent and consistent.
iii. Ifthe lines are parallelthen the pair ofequations has no solution. In this case, the pair
ofequationsisinconsistent.
4. We have discussed the following methods for finding the solution(s) of a pair of linear
equations.
i. ModelMethod. ii. GraphicalMethod
iii. Algebraic methods- Substitutionmethod andEliminationmethod.
5. There exists a relationbetween the coefficientsand nature of systemof equations.
i. If
1 1
2 2
a b
a b
¹ thenthe pair oflinear equations is consistent.
ii. If 1 1 1
2 2 2
=
a b c
a b c
¹ then the pair oflinearequationsisinconsistent.
iii. If 1 1 1
2 2 2
= =
a b c
a b c
then the pair oflinear equations is dependent and consistent.
6. There are several situations which canbe mathematicallyrepresented bytwo equations
that are not linear to start with. But we canalter themso that they willbe reduced to a pair
oflinearequations.
Suggested Projects
l Construct somepairs oflinear equations fromdailylife situations and find solutionsofthe
equations byusing graphs.

5.1 INTRODUCTION
Sports committee of Dhannur High School wants to construct a Kho-Kho court of
dimensions 29 m × 16 m. This is to be
laid ina rectangular plot ofarea 558 m2
.
Theywant to leave space ofequalwidth
all around the court for the spectators.
What would be the width of the space
for spectators?Would it be enough?
Suppose the widthofthe space be
x meter.So, fromthe figurethe lengthof
the plot would be (29 + 2x) meter.
And, breadthofthe rectangular plot would be = (16 + 2x) m.
Therefore, area ofthe rectangular plot = length× breadth
= (29 + 2x) (16 + 2x)
Since the area ofthe plot is = 558 m2
(29 + 2x) (16 + 2x) = 558
4x2
+ 90x + 464 = 558
4x2
+ 90x - 94 = 0 (dividing by2)
2x2
+ 45x - 47 = 0
2x2
+ 45 x - 47 = 0 ..... (1)
Inprevious classwe solved the linear equations ofthe formax + b= c to find the value of
‘x’. Similarly, thevalue ofx from theabove equation willgivethe possible widthofthe space for
spectators.
Can you think of more such examples where we have to find the quantities, like in the
above example andget such equations.
Let usconsider another example:
Ranihas asquare metalsheet. Sheremoved squares ofside 9 cmfrom eachcorner ofthis
sheet. Of the remaining sheet, she turned up the sides to form an open box as shown. The
capacityofthe box is 144 cm3
. Can we find out the dimensions ofthe metalsheet?
Quadratic Equations
5
29 .
m
16
.
m
x
x
29+2 .
x m
16+2
.
x
m

Class-X Mathematics
106
Suppose the side of the square piece of metal sheet
be ‘x’ cm.
Then, the dimensionsofthe boxare
9 cm × (x-18) cm × (x-18) cm
Since volume of the box is 144 cm3
9 (x-18) (x-18) = 144
(x-18)2
= 16
x2
- 36x + 308 = 0
So, the side ‘x’ ofthe metalsheet have to satisfythe
equation.
x2
- 36x + 308 = 0 ..... (2)
Let us observe the L.H.S of equation(1) and (2)
Are theyquadratic polynomials?
We studied quadratic polynomials ofthe form ax2
+ bx + c,
a ¹ 0 in the previous chapter.
Since, the LHS ofthe aboveequations are quadratic polynomials and the RHS is 0 they
are called quadratic equations.
Inthis chapterwe willstudyquadratic equations and methods to find their roots.
5.2 QUADRATIC EQUATIONS
A quadratic equationin the variable x is an equationof the formax2
+ bx + c = 0, where
a, b, c are real numbers and a ¹ 0. For example, 2x2
+ x - 300 = 0 is a quadratic equation,
Similarly, 2x2
- 3x + 1 = 0, 4x - 3x2
+ 2 = 0 and 1 - x2
+ 300 = 0 are also quadratic equations.
In fact, any equation of the form p(x) = 0, where p(x) is a polynomial of degree 2, is a
quadraticequation. Whenwe write theterms of p(x) indescending order of their degrees, then
we get the standard formof theequation. That is, ax2
+ bx + c = 0, a ¹ 0 is called the standard
formofa quadratic equation and y = ax2
+ bx + c is called a quadratic function.
TRY THIS
Check whether thefollowing equations are quadratic or not ?
(i) x2
- 6x - 4 = 0 (ii) x3
- 6x2
+ 2x - 1 = 0
(iii) 7x = 2x2
(iv) 2
2
1
2
x
x
+ = (x ¹ 0)
(v) (2x + 1) (3x + 1) = b(x - 1) (x - 2) (vi) 3y2
= 192
9 cm.
9 cm.
9 cm. 9 cm.
x cm.
x
cm.
9 cm.
x - 18
x
-
1
8

Quadratic Equations 107
There are various situationsdescribed byquadratic functions. Some ofthemare:-
1. When a rocket isfired upward, thenthe path of the rocket isdefined by
a ‘quadratic function.’
2. Shapes ofthe satellite dish, reflecting mirror in a telescope, lens ofthe
eyeglassesandorbitsofthecelestialobjectsaredefinedbythequadratic
functions.
SatelliteDish Reflecting Mirror Lens of Spectacles
3. The path of a projectile is defined by a quadratic
function.
4. When the brakes are applied to a vehicle, the stopping distance is calculated byusing
a quadratic equation.
Example-1. Represent thefollowing situations withsuitablemathematicalequations.
i. Sridhar and Rajendar together have 45 marbles. Both of them lost 5 marbles each,
and the product ofthe number ofmarbles now theyhave is124. We would like to find
out how manymarbles each ofthemhad previously.
ii. The hypotenuse of a right triangle is 25 cm. Weknow that the difference in lengths of
the other two sides is 5 cm. We would like to find out the length ofthe two sides?
Solution : i. Let the number of marbles Sridhar had be x.
Earth
Sun
Earth
Artificial
Satellite

Class-X Mathematics
108
Then the number of marbles Rajendar had = 45 – x (Why?).
The number of marbles left with Sridhar, when he lost 5 marbles = x – 5
The number of marbles left withRajendar, when he lost 5 marbles = (45 – x) – 5
= 40 – x
Therefore, their product = (x – 5) (40 – x)
= 40x – x2
– 200 + 5x
= – x2
+ 45x – 200
So, – x2
+ 45x – 200 = 124 (Given that product = 124)
i.e.,– x2
+ 45x – 324 = 0
i.e., x2
– 45x + 324 = 0 (Multiplyby-1)
Therefore, thenumber ofmarbles Sridharhad‘x’,should satisfythequadraticequation
x2
– 45x + 324 = 0
which isthe required representationofthe problem.
ii. Let the length ofsmaller side be xcm
Then length oflarger side = (x + 5) cm
Given length ofhypotenuse = 25 cm
We know that ina right angle triangle (hypotenuse)2
= (side)2
+ (side)2
So, x2
+ (x + 5)2
= (25)2
x2
+ x2
+ 10x + 25 = 625
2x2
+ 10x - 600 = 0
x2
+ 5x - 300 = 0
Value of x fromthe above equation willgive the possible value of length of sides of the
givenright angledtriangle.
Example-2. Check whether thefollowing are quadratic equations:
i. (x – 2)2
+ 1 = 2x – 3 ii. x(x + 1) + 8 = (x + 2) (x – 2)
iii. x (2x + 3) = x2
+ 1 iv. (x + 2)3
= x3
– 4
Solution : i. LHS = (x – 2)2
+ 1 = x2
– 4x + 4 + 1 = x2
– 4x + 5
Therefore, (x – 2)2
+ 1 = 2x – 3 can be written as
x2
– 4x + 5 = 2x – 3
x
x
cm
+5
.
25
.
cm

i.e., x2
– 6x + 8 = 0
It is in the form of ax2
+ bx + c = 0.
Therefore, the given equation is aquadratic equation.
ii. Here LHS = x(x + 1) + 8 = x2
+ x + 8
and RHS = (x + 2)(x – 2) = x2
– 4
Therefore, x2
+ x + 8 = x2
– 4
x2
+ x + 8 - x2
+ 4 = 0
i.e., x + 12 = 0
It is not in the formof ax2
+ bx + c = 0, (a ¹ 0)
Therefore, the given equationis not a quadratic equation.
iii. Here, LHS = x (2x + 3) = 2x2
+ 3x
So, x (2x + 3) = x2
+ 1 can be rewritten as
2x2
+ 3x = x2
+ 1
Therefore, we get x2
+ 3x – 1 = 0
+ bx + c = 0.
So, the given equation is a quadratic equation.
iv. Here, LHS = (x + 2)3
= (x + 2)2
(x + 2)
= (x2
+ 4x + 4) (x + 2)
= x3
+ 2x2
+ 4x2
+ 8x + 4x + 8
= x3
+ 6x2
+ 12x + 8
Therefore, (x + 2)3
= x3
– 4 can be rewritten as
x3
+ 6x2
+ 12x + 8 = x3
– 4
i.e.,6x2
+ 12x + 12 = 0 or, x2
+ 2x + 2 = 0
+ bx + c = 0.
So, the given equation is a quadratic equation.
Remark : In (ii) above, the given equation appears to be a quadratic equation, but it is not a
quadratic equation.
In (iv) above, the given equationappears to be a cubic equation (anequationofdegree
3) andnot aquadraticequation.But it turnsoutto beaquadraticequation.Asyoucansee,often
we needto simplifythe givenequationbefore deciding whetherit is quadraticor not.

Class-X Mathematics
110
EXERCISE - 5.1
1. Check whether the following are quadratic equations :
i. (x + 1)2
= 2(x – 3) ii. x2
– 2x = (–2) (3 – x)
iii. (x – 2)(x + 1) = (x – 1)(x + 3) iv. (x – 3)(2x +1) = x(x + 5)
v. (2x – 1)(x – 3) = (x + 5)(x – 1) vi. x2
+ 3x + 1 = (x – 2)2
vii. (x + 2)3
= 2x (x2
– 1) viii. x3
– 4x2
– x + 1 = (x – 2)3
2. Represent the following situations inthe formofquadratic equations :
i. The areaofa rectangular plot is 528 m2
. Thelength ofthe plot is one metremore than
twice its breadth. We need to find the length and breadthof the plot.
ii. The product oftwo consecutive positive integers is 306. We need to find the integers.
iii. Rohan’s mother is 26 years older than him.The product oftheir ages after 3 years will
be 360 years. We need to find Rohan’s present age.
iv. A train travels a distance of480 kmat a uniformspeed. Ifthe speed had been 8 km/h
less, then it would have taken 3 hours more to cover the same distance. We need to
find the speed ofthe train.
5.3 SOLUTION OF A QUADRATIC EQUATION BY FACTORISATION
We have learned to represent some of the daily life situations in the form of quadratic
equationwith anunknownvariable ‘x’.
Now we need to find the value of x.
Consider the quadratic equation 2x2
– 3x + 1 = 0. If we replace x by 1. Then, we get
(2 × 12
) – (3 × 1) + 1 = 0 = RHS of the equation. Since 1 satisfies the equation , we saythat 1
is a root of the quadratic equation 2x2
– 3x + 1 = 0.
x = 1 is a solution ofthe quadratic equation.
This also means that 1 is a zero ofthe quadratic polynomial2x2
– 3x + 1.
In general, a real number a is called a root of the quadratic equation ax2
+ bx + c = 0,
if aa2
+ b a + c = 0. We also say that x = a is a solution of the quadratic equation, or
a satisfies the quadratic equation.
Note that the zeroes of the quadratic polynomial ax2
+ bx + c (a ¹ 0) and the roots
of the quadratic equation ax2
+ bx + c = 0 (a ¹ 0) are the same.
We have observed, inChapter 3, that a quadratic polynomialcanhave at most two zeroes.
So, anyquadratic equation can have at most two roots. (Why?)

We have learntinClass-IX, howto factorise quadraticpolynomialsbysplitting theirmiddle
terms. We shalluse this knowledge for finding the roots ofa quadratic equation. Let us see.
Example-3. Find the roots of the equation 2x2
– 5x + 3 = 0, byfactorisation.
Solution : Let us first split the middle term. Recall that if ax2
+ bx + c is a quadratic
polynomial then to split the middle termwe have to find two numbers p and q suchthat p + q=
b and p × q = a × c. So to split the middle termof 2x2
– 5x + 3, we have to find two numbers
p and q such that p + q = –5 and p × q = 2 × 3 = 6.
For this we have to list out all possible pairs of factors of 6. They are (1, 6), (–1, –6);
(2, 3); (–2, –3). From the list it is clear that the pair (–2, –3) will satisfy our condition
p + q = –5 and p × q = 6.
The middle term ‘–5x’ can be written as ‘–2x – 3x’.
So, 2x2
– 5x + 3 = 2x2
– 2x – 3x + 3 = 2x (x – 1) –3(x – 1) = (2x – 3)(x – 1)
Now, 2x2
– 5x + 3 = 0 can be rewritten as (2x – 3)(x – 1) = 0.
So, the values of x for 2x2
– 5x + 3 = 0 are the same for (2x – 3)(x – 1) = 0,
i.e., either 2x – 3 = 0 or x – 1 = 0.
Now, 2x – 3 = 0 gives x =
3
2
and x – 1 = 0 gives x = 1.
So, x =
3
2
and x = 1 are the solutions ofthe equation.
In other words, 1 and
3
2
are the roots of the equation 2x2
– 5x + 3 = 0.
Do This
Find the rootsofthe following equationsusing factorisationmethod.
(i) x2
+ 5x + 6 = 0 (ii) x2
- 5x + 6= 0
(iii) x2
+ 5x - 6 = 0 (iv) x2
- 5x – 6= 0
TRY THIS
Verify whether 1 and
3
2
are the roots of the equation 2x2
– 5x + 3 = 0.
Note that we have found the roots of 2x2
– 5x + 3 = 0 by factorising 2x2
– 5x + 3
into two linear factors and equating each factor to zero.

Class-X Mathematics
112
Example 4 : Find the roots of the equation x-
1
3x
=
1
6
(x ¹ 0)
Solution : We have x-
1
3x
=
1
6
Þ 6x2
- x - 2 = 0
6x2
– x – 2 = 6x2
+ 3x – 4x – 2
= 3x (2x + 1) – 2 (2x + 1)
= (3x – 2)(2x + 1)
The roots of 6x2
– x – 2 = 0 are the values of x for which (3x – 2)(2x + 1) = 0
Therefore, 3x – 2 = 0 or 2x + 1 = 0,
i.e.,x =
2
3
or x =
1
2
-
Therefore, the roots of 6x2
– x – 2 = 0 are
2
3
and
1
2
- .
We verify the roots, by checking that
2
3
and
1
2
- satisfy 6x2
– x – 2 = 0.
Example-5. Find the width ofthe space for spectators discussed insection 5.1.
Solution : In Section5.1, we found that if the width ofthe space for spectators is x m, then x
satisfiesthe equation2x2
+45x -47=0.Applyingthefactorisationmethodwewritethisequation
as:-
2x2
- 2x + 47x - 47 = 0
2x (x - 1) + 47 (x - 1) = 0
i.e., (x - 1) (2x + 47) = 0
So, the roots ofthe given equation are x = 1or x =
47
2
-
. Since ‘x’ is the width of space
of the spectators it cannot be negative.
Thus, the widthis 1 m. So it is not enoughfor spectators.
EXERCISE - 5.2
1. Find the rootsofthe following quadraticequations byfactorisation:
i. x2
– 3x – 10 = 0 ii. 2x2
+ x – 6 = 0 iii. 2
2 7 5 2 0
+ + =
x x
iv.
2 1
2 0
8
- + =
x x v. 100x2
– 20x + 1 = 0 vi. x(x + 4) = 12
vii. 3x2
– 5x + 2 = 0 viii.
3
2
- =
x
x
(x ¹ 0) ix. 3(x – 4)2
– 5(x – 4) = 12

2. Find two numbers whose sumis 27 and product is 182.
3. Find two consecutive positive integers, sumofwhose squares is 613.
4. The altitude ofaright triangleis7 cmlessthanitsbase. Ifthe hypotenuse is13 cm, find the
other two sides.
5. Acottage industry produces a certainnumber of potteryarticles in a day. It was observed
on a particular daythat the cost ofproduction ofeach article (in rupees) was3 more than
twice thenumber ofarticles produced on that day. Ifthe totalcost ofproduction on that
daywas Rs 90, find the number ofarticles produced and the cost ofeach article.
6. Find the dimensions of a rectangle whose perimeter is 28 meters and whose area is 40
square meters.
7. The baseofa triangle is4cmlonger thanitsaltitude. Ifthe area ofthe triangleis48 sq.cm
thenfind its base and altitude.
8. Two trainsleavearailwaystationatthesametime. Thefirst traintravelstowardswest and
the second train towardsnorth. The first traintravels5 km/hr faster thanthe second train.
Ifafter two hours they are 50 km. apart, find the average speed ofeach train.
9. In a class of 60 students, each boy contributed rupees equal to the number of girls and
each girlcontributed rupees equalto the number ofboys. Ifthe totalmoneythen collected
was D1600. How manyboys were there in the class?
10. A motor boat heads upstreama distance of24 kmin a river whose current is running at 3
kmperhour. The trip upandbacktakes6hours. Assuming that themotorboat maintained
a constant speed, what was its speed in stillwater?
5.4 SOLUTION OF A QUADRATIC EQUATION BY COMPLETING THE SQUARE
Intheprevious section, we havelearnt method offactorisationfor obtaining the rootsofa
quadratic equation. Is method offactorization applicable to alltypes ofquadratic equation? Let
us tryto solve x2
+ 4x - 4 = 0 byfactorisation method
To solve the given equation x2
+ 4x - 4 = 0 byfactorization method.
We have to find ‘p’and ‘q’ such that p + q = 4 and
p × q = -4
Wehaveno integersp, qsatisfyingaboveequation.So byfactorizationmethod it isdifficult
to solvethe givenequation.
Therefore, we shalltryanother method.

Class-X Mathematics
114
Consider thefollowing situation
The product ofSunita’s age (in years) two years ago and her agefour years hence is one
more than twice her present age. What is her present age?
To answer this, let her present age be x years. Their age two years ago would be x– 2 and
the age after four years will be x + 4. So, the product of both the ages is (x – 2)(x + 4).
Therefore, (x – 2)(x + 4) = 2x + 1
i.e., x2
+ 2x – 8 = 2x + 1
i.e., x2
– 9 = 0
So, Sunita’s present age satisfies the quadratic equation x2
– 9 = 0.
We can write this as x2
= 9. Taking square roots, we get x = 3 or x = – 3. Since the age is
a positive number, x = 3.
So, Sunita’s present age is 3 years.
Now consider another quadratic equation (x + 2)2
– 9 = 0. To solve it, we can write it as
(x + 2)2
= 9. Taking square roots, we get x + 2 = 3 or x + 2 = – 3.
Therefore, x = 1 or x = –5
So, the roots of the equation (x + 2)2
– 9 = 0 are 1 and – 5.
In both the examples above, the term containing x is inside a square, and we found the
roots easilybytaking the square roots.But, what happens ifwe are asked to solve the equation
x2
+ 4x – 4 = 0, which cannot be solved by factorisation also.
So, we now introduce the method of completing the square. The idea behind this
method is to adjust the left side of the quadratic equation so that it becomes a perfect
square of the first degree polynomial and the RHS without x term.
The process is as follows:
x2
+ 4x – 4 = 0
Þ x2
+ 4x = 4
x2
+ 2. x . 2 = 4
Now, the LHS isin the formof a2
+ 2ab. Ifwe add b2
it becomes as a2
+ 2ab + b2
which
is perfect square. So, byadding b2
= 22
= 4 to both sideswe get,
x2
+ 2.x.2 + 22
= 4 + 4
Þ (x + 2)2
= 8 Þ x + 2 = 8
±
Þ x = –2 2 2
±

Now consider the equation 3x2
– 5x + 2 =0. Note that the coefficient of x2
is not 1. So
we divide the entire equation by 3 so that the coefficient of x2
is 1

2 5 2
0
3 3
- + =
x x
Þ
2 5 2
3 3
-
- =
x x
Þ
2 5 2
2. .
6 3
-
- =
x x
Þ
2 2
2 5 5 2 5
2. .
6 6 3 6
-
æ ö æ ö
- + = +
ç ÷ ç ÷
è ø è ø
x x
2
5
add both side
6
æ ö
æ ö
ç ÷
ç ÷
ç ÷
è ø
è ø
2
5 2 25
6 3 36
-
æ ö
- = +
ç ÷
è ø
x
( ) ( )
2
12 2 25 1
5
=
6 36
x
´ - + ´
æ ö
-
ç ÷
è ø
2
5 24 25
=
6 36
x
- +
æ ö
-
ç ÷
è ø
2
5 1
6 36
æ ö
- =
ç ÷
è ø
x (take square root both sides)
5 1
6 6
- = ±
x
So,
5 1
6 6
= +
x or
5 1
6 6
= -
x
Therefore, x = 1 or x =
4
6
i.e., x = 1 or x =
2
3
Therefore, the roots ofthe given equationare 1 and
2
3
.
Fromtheaboveexampleswecandeducethefollowingalgorithmforcompletingthe square.
Algorithm : Let the quadratic equation be ax2
+ bx + c = 0 (a ¹ 0)
Step-1 : Divide each side by‘a’

Class-X Mathematics
116
Step-2 : Rearrange the equationso that constant termc/a ison the right side. (RHS)
Step-3 :Add
2
1
2
é ù
æ ö
ç ÷
ê ú
è ø
ë û
b
a
to both sides to make LHS, a perfect square.
Step-4 : Write the LHS asa square and simplifythe RHS.
Step-5 : Solve it.
Example-6. Find the roots of the equation 5x2
– 6x – 2 = 0 by the method of completing the
square.
Solution : Given : 5x2
– 6x – 2 = 0
Now wefollow theAlogarithm
Step-1 :
2 6 2
0
5 5
- - =
x x (Dividing bothsides by5)
Step-2 :
2 6 2
5 5
- =
x x
Step-3 :
2 2
2 6 3 2 3
5 5 5 5
æ ö æ ö
- + = +
ç ÷ ç ÷
è ø è ø
x x
2
3
Adding to both sides
5
æ ö
æ ö
ç ÷
ç ÷
ç ÷
è ø
è ø
Step-4 :
2
3 2 9
5 5 25
æ ö
- = +
ç ÷
è ø
x
Step-5 :
2
3 19
5 25
æ ö
- =
ç ÷
è ø
x
3 19
5 25
- = ±
x
3 19
5 5
= +
x or
3 19
5 5
= -
x

3 19
5
+
=
x or
3 19
5
-
=
x

Example-7. Find the roots of 4x2
+ 3x + 5 = 0 by the method ofcompleting the square.
Solution : Given 4x2
+ 3x + 5 = 0
2 3 5
0
4 4
+ + =
x x
2 3 5
4 4
-
+ =
x x
2 2
2 3 3 5 3
4 8 4 8
-
æ ö æ ö
+ + = +
ç ÷ ç ÷
è ø è ø
x x
2
3 5 9
8 4 64
-
æ ö
+ = +
ç ÷
è ø
x
2
3 71
0
8 64
-
æ ö
+ = <
ç ÷
è ø
x
But
2
3
8
æ ö
+
ç ÷
è ø
x cannot be negative foranyrealvalue ofx (Why?). So, there is no real value
ofx satisfying thegiven equation. Therefore, thegiven equation hasno real roots.
DO THIS
Solve the equations bycompleting the square
(i) x2
- 10x + 9 = 0 (ii) x2
- 5x + 5= 0
(iii) x2
+ 7x - 6 = 0
We have solved severalexamples with the use ofthe method of‘completing the square.’
Now, let us applythis method in standard formofquadratic equation ax2
+bx+c=0 (a¹0).
Step 1 : Dividing the equationby‘a’we get
2
0
+ + =
b c
x x
a a
Step 2 : 2
+ = -
b c
x x
a a

Class-X Mathematics
118
Step 3 :
2 2
2 1 1
2 2
é ù é ù
+ + = - +
ê ú ê ú
ë û ë û
b b c b
x x
a a a a
2
1
adding both sides
2
b
a
é ù
é ù
ê ú
ê ú
ë û
ê ú
ë û
Q
2 2
2
2
2 2 2
é ù é ù
Þ + × + = - +
ê ú ê ú
ë û ë û
b b c b
x x
a a a a
Step 4 :
2 2
2
4
2 4
-
é ù
+ =
ê ú
ë û
b b ac
x
a a
Step 5 : If b2
- 4ac > 0, then by taking the square roots, we get
2
4
2 2
± -
+ =
b b ac
x
a a
Therefore,
2
4
2
- ± -
=
b b ac
x
a
So, the roots of ax2
+ bx + c = 0 are
2
4
2
- + -
b b ac
a
and
2
4
2
- - -
b b ac
a
,
if b2
– 4ac > 0.
If b2
– 4ac < 0, the equation will have no realroots. (Why?)
Thus, if b2
– 4ac > 0, then the roots of the quadratic equation ax2
+ bx + c = 0 are
given by
2
4
2
b b ac
a
- ± -
.
Thisformulaforfindingtherootsofaquadraticequationisknownasthequadraticformula.
Let us consider some examples byusing quadratic formula.
Example-8. Solve Q. 2(i)of Exercise 5.1 byusing the quadratic formula.
Solution : Let the breadth ofthe plot be x metres.
Then the length is (2x + 1) metres.
Since area of rectangular plot is 528 m2
,
we can write x(2x + 1) = 528, i.e., 2x2
+ x – 528 = 0.
This is in the form of ax2
+ bx + c = 0, where a = 2, b = 1, c = – 528.
So, the quadratic formula gives usthe solution as

1 1 4(2)(528) 1 4225 1 65
4 4 4
- ± + - ± - ±
= = =
x
i.e.,
64
4
=
x or
66
4
-
=
x
i.e., x = 16 or x =
33
2
-
Since x cannot be negative. So, the breadthofthe plot is 16 metres and hence, the length
of the plot is (2x + 1) = 33m.
You shouldverifythat these valuessatisfythe conditions ofthe problem.
THINK - DISCUSS
We have three methods to solve a quadratic equation. Among these three, which
method would you like to use? Why?
Example-9. Find two consecutive positive odd integers, sumofwhose squaresis 290.
Solution: Let thefirst positiveodd integer be x.Then,thesecondintegerwillbex +2.According
to the question,
x2
+ (x + 2)2
= 290
i.e., x2
+ x2
+ 4x + 4 = 290
i.e., 2x2
+ 4x – 286 = 0
i.e., x2
+ 2x – 143 = 0
which is a quadratic equation in x.
Using thequadratic formula
2
4
=
2
b b ac
x
- ± -
we get,
2 4 572 2 576 2 24
2 2 2
- ± + - ± - ±
= = =
x
i.e., x = 11 or x = – 13
But x is given to be positive odd integer. Therefore, x ¹ – 13.
Thus, the two consecutive odd integers are 11 and (x + 2) = 11 + 2 = 13.
Check : 112
+ 132
= 121 + 169 = 290.

Class-X Mathematics
120
Example-10. Arectangular parkis to be designed whose breadthis 3 mlessthan its length. Its
area is to be 4 square metres more than the area of a park that has already been made in the
shape ofanisosceles triangle with its base as the breadthofthe rectangular park and ofaltitude
12 m. Find its length and breadth.
Solution : Let the breadth ofthe rectangular park be x m.
So, its length= (x + 3) m.
Therefore, the area ofthe rectangular park = x(x + 3) m2
= (x2
+ 3x) m2
.
Now, base ofthe isosceles triangle = x m.
Therefore, its area =
1
2
× x × 12 = 6 x m2
.
Accordingtoour requirements,
x2
+ 3x = 6x + 4
i.e., x2
– 3x – 4 = 0
Using the quadraticformula, we get
3 25 3 5
2 2
± ±
= =
x = = 4 or – 1
But x ¹ – 1 (Why?). Therefore, x = 4.
So, the breadth of the park = 4m and its length will be x + 3 = 4 + 3 = 7m.
Verification : Area of rectangular park = 28 m2
,
area of triangular park = (28 – 4)m2
= 24m2.
Example-11. Find the roots ofthe following quadraticequations, iftheyexist.
(i) x2
+ 4x + 5 = 0 (ii) 2x2
– 2 2 x + 1 = 0
Solution :
(i) x2
+ 4x + 5 = 0. Here, a = 1, b = 4, c = 5. So, b2
– 4ac = 16 – 20 = – 4 < 0.
Since the square ofa real number cannot be negative, therefore 2
4
-
b ac will not
have anyreal value.
So, there are no realroots for the given equation.
(ii) 2x2
– 2 2 x + 1 = 0. Here, a = 2, b = -2 2 , c = 1.
12
x
+
3
x

So, b2
– 4ac = 8 – 8 = 0
Therefore,
2 2 0 2
0
4 2
±
= = ±
x i.e., x =
1
2
.
So, the roots are
1 1
, .
2 2
Example-12. Find the roots ofthe following equations:
(i)
1
3, 0
+ = ¹
x x
x
(ii)
1 1
3, 0,2
2
- = ¹
-
x
x x
Solution :
(i)
1
3
+ =
x
x
. Multiplying both sides of equation byx, we get
x2
+ 1 = 3x
i.e., x2
– 3x + 1 = 0, which is a quadratic equation.
Here, a = 1, b = – 3, c = 1
So, b2
– 4ac = 9 – 4 = 5 > 0
Therefore,
3 5
2
±
=
x (why?)
So, the roots are
3 5
2
+
and
3 5
.
2
-
(ii)
1 1
3, 0, 2.
2
- = ¹
-
x
x x
As x ¹ 0, 2, multiplying the equation by x (x – 2), we get
(x – 2) – x = 3x (x – 2)
= 3x2
– 6x
So, the given equation reduces to 3x2
– 6x + 2 = 0, which is a quadratic equation.
Here, a = 3, b = – 6, c = 2. So, b2
– 4ac = 36 – 24 = 12 > 0
Therefore,
6 12 6 2 3 3 3
.
6 6 3
± ± ±
= = =
x

Class-X Mathematics
122
So, the roots are
3 3
3
+
and
3 3
.
3
-
Example-13. Amotor boat whose speed is 18 km/h instill water. It takes 1 hour more to go
24 kmupstreamthan to return downstreamto the same spot. Find the speed of the stream.
Solution : Let the speed ofthe stream be x km/h.
Therefore, the speed of the boat upstream = (18 – x) km/h and the speed of the boat
downstream = (18 + x) km/h.
The time takento go upstream =
distance
speed
=
24
18 x
-
hours.
Similarly, thetime taken to go downstream=
24
18+ x
hours.
According tothe question,
24 24
1
18 18
- =
- +
x x
i.e., 24(18 + x) – 24(18 – x) = (18 – x) (18 + x)
i.e., x2
+ 48x – 324 = 0
Using the quadraticformula, we get
2
48 48 1296 48 3600
2 2
- ± + - ±
= =
x
48 60
6
2
- ±
= = or -54
Since x is the speed ofthe stream, it cannot be negative. So, we ignore the root x = – 54.
Therefore, x = 6 gives the speed of the streamas 6 km/h.
EXERCISE - 5.3
1. Find the roots ofthefollowing quadraticequations, iftheyexist.
i. 2x2
+ x – 4 = 0 ii. 2
4 4 3 3 0
+ + =
x x
iii. 5x2
- 7x - 6 = 0 iv. x2
+ 5 = -6x

2. Find the roots of the quadratic equations given in Q.1 byapplying the quadratic formula.
3. Find the rootsofthefollowing equations:
(i)
1
3, 0
- = ¹
x x
x
(ii)
1 1 11
, 4,7
4 7 30
- = ¹ -
+ -
x
x x
4. The sumofthe reciprocals ofRehman’sages, (inyears) 3 years ago and 5 years fromnow
is
1
3
. Find his present age.
5. Ina class test, thesumofMoulika’s marks in Mathematics and English is 30. Ifshe got 2
marks morein Mathematicsand 3 marks less in English, the product ofher marks would
have been 210. Find her marks inthe two subjects.
6. The diagonalof a rectangular field is 60 metres more than the shorter side. If the longer
side is 30metres more than the shorter side, find the sidesofthe field.
7. The difference of squares of two numbers is 180. The square ofthe smaller number is 8
times the larger number. Find thetwo numbers.
8. A train travels 360 km at a uniform speed. If the speed had been 5 km/h more, it would
have taken 1 hour less for the same journey. Find the speed ofthe train.
9. Two water taps together can filla tank in
3
9
8
hours. The tap oflarger diameter takes 10
hours less thanthe smaller one to fillthe tank separately. Find the time inwhicheachtap
canseparatelyfillthe tank.
10. An express train takes 1 hour less than a passenger train to travel132 kmbetween Mysore
andBangaluru(withouttakinginto considerationthetimetheystopat intermediatestations).
Ifthe average speed of the expresstrain is 11km/h more than that ofthe passenger train,
find the average speed ofthe two trains.
11. Sumoftheareas oftwo squares is 468 m2
. Ifthe difference oftheir perimetersis24m, find
the sides ofthe two squares.
12. Anobject isthrownupwards withaninitialvelocityof17m/sec froma buildingwith12 m
height. It isat a height of S = 12 + 17t – 5t2
fromthe ground after a flight of‘t’seconds.
Find the time taken bythe object to touch the ground.
13. If a polygonof‘n’ sides has
1
2
n (n-3) diagonals. How manysides are there ina polygon
with 65 diagonals? Is there a polygonwith 50 diagonals?

Class-X Mathematics
124
5.5 NATURE OF ROOTS
In the previous section, we have seenthat the roots of the equation ax2
+ bx + c = 0 are
givenby
2
4
2
- ± -
=
b b ac
x
a
Now let ustryto studythe nature ofroots.
Remember that zeroes are those points where value of polynomialbecomes zero or we
cansaythat the curve ofquadraticpolynomialcuts the X-axis.
Similarly, roots ofa quadratic equation arethosepoints where the curve cuts the X-axis.
Case-1 : If b2
- 4ac > 0;
We get two distinct realroots
2
4
2
b b ac
a
- + -
,
2
4
2
b b ac
a
- - -
Insuchcase ifwe draw corresponding graphfor the given quadratic equation we get the
followingtypesoffigures.
Figure shows that the corresponding curve of the quadratic equation cuts the X-axis at two
distinct points
Case-2 : If b2
- 4ac = 0
x =
0
2
b
a
- +
So, x =
2
b
a
-
,
2
-b
a
Figure shows that the graph ofthe quadratic equation touchesX-axisat one point.
Case-3 : b2
- 4ac < 0
There are no realroots. Roots are imaginary.

In this case, the graph neither intersects nor touches the X-axisat all. So, there are no real
roots.
Since b2
– 4ac determines whether the quadratic equation ax2
+ bx + c = 0 (a ¹ 0) has
realroots or not, b2
– 4ac is called the discriminant ofthe quadratic equation.
So, a quadratic equation ax2
+ bx + c = 0 (a ¹ 0) has
i. two distinct real roots, if b2
– 4ac > 0,
ii. two equalreal roots, if b2
– 4ac = 0,
iii. no real roots, if b2
– 4ac < 0.
Example-14. Find the discriminant ofthe quadratic equation 2x2
– 4x + 3 = 0, and hence find
the natureofits roots.
Solution : The given equation is in the form of ax2
+ bx + c = 0, where a = 2, b = – 4 and
c = 3. Therefore, the discriminant
b2
– 4ac = (– 4)2
– (4 × 2 × 3) = 16 – 24 = – 8 < 0
So, the given equation has no realroots.
Example-15. Apole hasto be erected at apoint on the boundaryof acircular park ofdiameter
13metresinsuchawaythat thedifferencesofitsdistancesfromtwo diametricallyoppositefixed
gatesAandB onthe boundaryis7 metres. Is it possible to do so?Ifyes, at what distances from
the two gates should the pole be erected?
Solution : Let us first draw the diagram.
Let P be therequired locationofthe pole. Let the distance ofthe
pole fromthe gate B be x m, i.e., BP = x m. Now the difference ofthe
distances ofthepole fromthe two gates=AP– BP(or, BP–AP)= 7m.
Therefore, AP = (x + 7) m.
Now,AB = 13m, and sinceAB is a diameter,
APB
Ð = 900
(Why?)
Therefore, AP2
+ PB2
= AB2
(ByPythagoras theorem)
i.e., (x + 7)2
+ x2
= 132
i.e., x2
+ 14x + 49 + x2
= 169
i.e., 2x2
+ 14x – 120 = 0
A
B
P
13 m

Class-X Mathematics
126
So, the distance ‘x’ ofthe pole fromgate B satisfies the equation
x2
+ 7x – 60 = 0
So, it would be possible to place the pole ifthis equationhas realroots. To see ifthis is so or not,
let usconsider its discriminant. Thediscriminant is
b2
– 4ac = 72
– 4 × 1 × (– 60) = 289 > 0.
So, thegivenquadratic equation has two realroots, and it is possible to erect the pole on
the boundaryofthe park.
Solving the quadratic equation x2
+ 7x – 60 = 0, bythe quadratic formula, we get
7 289 7 17
2 2
- ± - ±
= =
x
Therefore, x = 5 or – 12.
Since x is the distance between the pole and the gate B, it must be positive.
Therefore, x = – 12 will have to be ignored. So, x = 5.
Thus, the pole hasto be erected onthe boundaryofthe park at adistance of 5mfromthe
gate B and 12mfrom the gateA.
TRY THIS
1. Explainthebenefitsofevaluatingthediscriminantofaquadraticequationbeforeattempting
to solve it. What does itsvalue signifies?
2. Write three quadratic equations, one having two distinct realsolutions, one having no
realsolutionand one having exactlyone realsolution.
Example-16. Findthediscriminantoftheequation 2 1
3 2
3
- +
x x = 0 and hence find the nature
ofitsroots. Find them, iftheyare real.
Solution : Here a = 3, b = – 2 and c =
1
3
Therefore, discriminant 2 2 1
4 ( 2) 4 3 4 4 0.
3
- = - - ´ ´ = - =
b ac
Hence, the givenquadratic equationhas two equal realroots.
The roots are , ,
2 2
- -
b b
a a
i.e.,
2 2
, ,
6 6
i.e.,
1 1
, .
3 3

EXERCISE - 5.4
1. Find the nature ofthe roots ofthe following quadratic equations. If realroots exist, find
them:
(i) 2x2
– 3x + 5 = 0 (ii) 2
3 4 3 4 0
- + =
x x
(iii) 2x2
– 6x + 3 = 0
2. Find the values ofk for each of the following quadratic equations, so that theyhave two
equalroots.
(i) 2x2
+ kx + 3 = 0 (ii) kx (x – 2) + 6 = 0 (k ¹ 0)
3. Isit possibleto designa rectangular mango grove whose lengthistwiceitsbreadth, andthe
area is 800 m2
? Ifso, find its lengthand breadth.
4. The sumof the ages oftwo friends is 20 years. Four years ago, the product oftheir ages in
years was 48. Is the above situationpossible? Ifso, determine their present ages.
5. Is it possible to design a rectangular park ofperimeter 80 mand area 400 m2
? Ifso, find
its lengthand breadth. Comment onyou answer.
OPTIONAL EXERCISE
[For extensive Learning]
1. Some points are plotted on a plane such that any three of them are non collinear. Each
point isjoinedwith allremaining pointsbyline segments. Find the number of pointsifthe
number oflinesegments are 10.
2. A two digit number is such that the product of its digits is 8. When 18 is added to the
number theyinterchange their places. Determine the number.
3. A piece ofwire 8 m. in length is cut into two pieces, andeach piece is bent into a square.
Where should the cut in the wire be made ifthe sumof the areas of these squares is to be
2 m2
?
2 2 2 2
8
Hint : 8, 2 2
4 4 4 4
x y x x
x y
é ù
-
æ ö æ ö æ ö æ ö
+ = + = Þ + =
ê ú
ç ÷ ç ÷ ç ÷ ç ÷
è ø è ø è ø è ø
ê ú
ë û
.
4. Vinayand Praveenworking together can paint the exterior ofa house in 6 days. Vinayby
himselfcan complete the job in 5days less than Praveen. How long willit takeVinayto
complete the job.
5. Show that the sum ofroots of a quadratic equation ax2
+ bx + c = 0 (a ¹ 0) is
-b
a
.

Class-X Mathematics
128
6. Show that the product ofthe roots ofa quadratic equation ax2
+ bx + c = 0 (a¹ 0) is
c
a
.
7. Ifthe sumofthe fractionanditsreciprocalis 2
16
21
, find the fraction.
1. Standard form of quadratic equation in variable x is ax2
+ bx + c = 0, where a, b, c are
real numbers and a ¹ 0.
2. A real number a is said to be a root of the quadratic equation ax2
+ bx + c = 0, if
aa2
+ ba + c = 0. The zeroes of the quadratic polynomial ax2
+ bx + c and the roots of
the quadratic equation ax2
+ bx + c = 0 are the same.
3. If we can factorise ax2
+ bx + c, a¹ 0, into a product of two linear factors, then the roots
ofthe quadratic equation ax2
+ bx + c = 0 can be found byequating eachfactor to zero.
4. A quadratic equationcan also be solved bythe method ofcompleting the square.
5. Quadratic formula: The roots ofa quadratic equation ax2
+ bx + c = 0 (a ¹ 0) are given
by
2
4
,
2
- ± -
b b ac
a
provided b2
– 4ac > 0.
6. A quadratic equation ax2
+ bx + c = 0 (a ¹ 0) has
(i) two distinct realroots, if b2
– 4ac > 0,
(ii) two equalroots (i.e., coincident roots), if b2
– 4ac = 0, and
(iii) no real roots, if b2
– 4ac < 0.
Suggested Projects
Solving quadratic equations by geometrical methods.
l Take two or three quadratic equations of the form ax2
+ bx + c = 0, where a ¹ 0, for
different situations like a > 0, a < 0, b = 0 and solve thembygraphical methods.

6.1 INTRODUCTION
You might have observed that innature, manythingsfollowa certainpatternsuchasthe
petals of a sunflower, the cells of a honeycomb, the grains on a maize cob, the spirals on a
pineapple and on a pine cone etc.
Can you see a pattern in each of the above given example? We can see the natural
patterns have a repetition which is not progressive. The identical petals ofthe sunflower are
equidistantlygrown.Inahoneycombidenticalhexagonalshapedcellsarearrangedsymmetrically
around each hexagonal cell. Similarly, you can find out other natural patterns in spirals of
pineapple....
You canlook for some other patterns in nature. Someexamples are:
(i) List ofthelast digits (digits inunit place) takenfromthe values of4, 42
, 43
, 44
, 45
, 46
.....
is
4, 6, 4, 6, 4, 6, ......
(ii) Mary is doing problems on patterns as part ofpreparing for a bank exam. One ofthem
is “find thenext two terms inthe following pattern”.
1, 2, 4, 8, 10, 20, 22 .......
(iii) Usha applied for a job and got selected. She has been offered a job with a starting
monthlysalaryof D8000, withanannualincrement of D500. Her salary(in rupees) for
to 1st
, 2nd
, 3rd
... years will be 8000, 8500, 9000 ..... respectively.
(iv) The lengths ofthe rungs ofaladderdecrease uniformlyby2 cmfrombottomto top. The
bottom rung is 45 cm in length. The lengths (incm) of the 1st
, 2nd
, 3rd
, .... 8th
rung from
the bottom to the top are 45, 43, 41, 39, 37, 35, 33, 31 respectively.
Can you see anyrelationship between the terms in the pattern ofnumbers written above?
Patterngiven inexample (i) has a relationof two numbers one after the other i.e. 4 and
6arerepeating alternatively.
Progressions
6

Class-X Mathematics
130
Now tryto find out the patterninexample(ii). Inexamples(iii) and (iv),the relationship
betweenthenumbers ineachlist is constantlyprogressive. Inthe givenlist 8000, 8500, 9000, ....
each succeeding termis obtained byadding 500 to the preceding term.
Where as in 45, 43, 41, ..... each succeeding term is obtained by adding ‘-2’ to each
preceding term. Now we can see some more examples ofprogressive patterns.
(a) In a savings scheme, the amount becomes
5
4
timesofitselfafter 3 years.
The maturityamount (in Rupees) of aninvestment of D8000 after 3, 6, 9 and 12 years
willbe 10000, 12500, 15625, 19531.25 respectively.
(b) The number ofunit squares insquares with sides 1, 2, 3, .... units are respectively,
12
, 22
, 32
, ....
(c) Hema put Rs. 1000 into her daughter’s money box when she was one year old and
increased the amount byRs. 500 everyyear. The amount ofmoney(in Rs.) inthe boxon
her 1st
, 2nd
, 3rd
, 4th
........ birthday would be.
1000, 1500, 2000, 2500, ..... respectively.
(d) The fraction offirst, second, third ..... shaded regions of the squares in the following
figurewillberespectively.
1 1 1 1
, , , , ....
4 16 64 256

Progressions 131
(e) Apair ofrabbits aretoo young to produce in their first month. In thesecond, and every
subsequent month, theyproduce a new pair. Each new pair of rabbits produce a new
pairintheirsecondmonthandineverysubsequent month(seethefigurebelow).Assuming
no rabbit dies, the number of pairs of rabbits at the start of the 1st
, 2nd
, 3rd
, ....., 6th
month, respectivelyare :
1, 1, 2, 3, 5, 8
In the examples above, we observe some patterns. In some of them, we find that the
succeeding termsare obtained byaddinga fixed number andinother bymultiplyingwitha fixed
number. In another, we find that theyare squares ofconsecutive numbers and so on.
In this chapter, we shall discuss some of these patterns in which succeeding terms are
obtained byadding a fixed numberto the preceding termor multiplying the precedingtermbya
fixed number. These patterns are called as arithmetic and geometric progressions respectively.
We shallalso see how to find their nth
termandthesumofnconsecutivetermsforageneralvalue
of ‘n’ and use this knowledge in solving some dailylifeproblems.
History : Evidenceisfoundthat by400BCE, BabyloniansknewofArithmeticandgeometric
progressions. According to Boethins (570 CE), these progressions were known to early
Greek writers.Among the Indian mathematicians,Aryabhatta (470 CE) was the first to give
formulaforthesumofsquaresand cubesofnaturalnumbersinhisfamousworkAryabhatiyam
written around 499 C.E. He also gave the formula for finding the sum of n terms of an
Arithmetic Progressionstarting with pth
term. Indian mathematician Brahmagupta(598 C.E),
Mahavira (850 C.E) and Bhaskara (1114-1185 C.E) also considered the sums of squares
and cubes.
1
1
2
3
5
8

Class-X Mathematics
132
6.2 ARITHMETIC PROGRESSIONS
Consider the followinglistsofnumbers:
(i) 1, 2, 3, 4, . . . (ii) 100, 70, 40, 10, . . .
(iii) – 3, –2, –1, 0, . . . (iv) 3, 3, 3, 3, . . .
(v) –1.0, –1.5, –2.0, –2.5, . . .
Each ofthe numbers in the list iscalled a term.
Canyou writethenext termineachofthe listsabove?Ifso, how willyou writeit?Perhaps
byfollowing a pattern or rule, let us observe and write the rule.
In (i), each termis 1 more than the termpreceding it.
In (ii), eachtermis 30 less than the termpreceding it.
In (iii), eachtermis obtained byadding 1 to the termpreceding it.
In (iv), allthe terms inthe list are 3 , i.e., each termis obtainedbyadding (or subtracting) 0 to the
termpreceding it.
In (v), eachtermis obtained byadding – 0.5 to (i.e., subtracting 0.5 from) the termpreceding it.
In all the lists above, we can observe that successive terms are obtained by adding or
subtractingafixednumbertotheprecedingterms.SuchlistofnumbersissaidtoformanArithmetic
Progression (AP).
TRY THIS
(i) Which ofthese areArithmetic Progressions and why?
(a) 2, 3, 5, 7, 8, 10, 15, ...... (b) 2, 5, 7, 10, 12, 15, ......
(c) -1, -3, -5, -7, ......
(ii) Write 3moreArithmetic Progressions.
6.2.1 WHAT IS AN ARITHMETIC PROGRESSION?
We observe that an arithmetic progression is a list of numbers in which each term,
except the first term is obtained by adding a fixed number to the preceding term.
This fixed number is called the common difference oftheAP.
Let us denote the first termofanAPby a1, second termbya2, . . ., nth termbyan andthe
common difference by d. Then theAP becomes a1, a2, a3, . . ., an.
So, a2 – a1 = a3 – a2 = . . . = an – an – 1 = d.

Progressions 133
Let us see some more examples ofAP:
(a) Heights( incm) ofsomestudentsofaschoolstandinginaqueueinthemorningassembly
are 147 , 148, 149, . . ., 157.
(b) Minimumtemperatures ( indegree celsius ) recorded for a week, in the monthofJanuary
in acity, arranged in ascending order are
– 3.1, – 3.0, – 2.9, – 2.8, – 2.7, – 2.6, – 2.5
(c) The balance money( in D) after paying 5% of the totalloan of D1000 everymonth is
950, 900, 850, 800, . . ., 50.
(d) Cash prizes ( in D ) given by a school to the toppers of Classes I to XII are 200, 250,
300, 350, . . ., 750 respectively.
(e) Totalsavings (inD) after everymonth, for 10 months whenRs.50 aresaved eachmonth
are 50, 100, 150, 200, 250, 300, 350, 400, 450, 500.
THINK AND DISCUSS
1. Think how eachofthe list givenabove formanAP. Discuss withyour friends.
2. Find the common difference ofeachofthe above lists? Think when is it positive?
3. Write an arithmetic progression in whichthe common difference is a small positive
quantity.
4. Make anAPinwhich the common difference is big(large) positive quantity.
5. Make anAPin which the commondifference is negative.
General form ofanAP: AnAP can be written as
a, a + d, a + 2d, a + 3d, . . .
This is called general form of anA.P where ‘a’is the first term and ‘d’is the common
difference
For example in 1, 2, 3, 4, 5, ....
The first termis 1 and the common difference isalso 1.
In 2, 4, 6, 8, 10 ....., what is the first termand what is the common difference?
ACTIVITY
(i) Make the followingfigures withmatchsticks

Class-X Mathematics
134
(ii) Write downthenumber ofmatch sticks required for each figure.
(iii) Can you find a commondifference in members ofthe list?
(iv) Does the list ofthese numbers formanAP?
6.2.2 PARAMETERS OF ARITHMETIC PROGRESSIONS
Note that inexamples (a) to (e) above, in section 6.2.1 there are only afinite number of
terms. Such anAPis called a finiteAP. Also note that each of theseArithmetic Progressions
(APs) has a last term. TheAPs in examples (i) to (v) in the section 6.2, are not finiteAPs and so
theyare called infiniteArithmetic Progressions. SuchAPs are never ending and do not have
a last term.
DO THIS
Writethree examplesfor finiteAPand threefor infiniteAP.
Now, to know about anAP, what is the minimuminformation that you need?Isit enough
to know the first term? Or, isit enough to know onlythe common difference?
We can see that we need to know both – the first term a and the common difference d.
These two parametersare sufficient forus to complete theArithmetic Progression.
For instance, ifthe first term a is 6 and the common difference d is 3, then theAPis
6, 9,12, 15, . . .
and if a is 6 and d is – 3, then theAP is
6, 3, 0, –3, . . .
Similarly,when
a = – 7, d = – 2, the AP is – 7, – 9, – 11, – 13, . . .
a = 1.0, d = 0.1, the AP is 1.0, 1.1, 1.2, 1.3, . . .
a = 0, d = 1
1
2
, the AP is 0, 1
1
2
, 3, 4
1
2
, 6, . . .
a = 2, d = 0, the AP is 2, 2, 2, 2, . . .
So, ifyou know what a and d are, you canlist theAP.
Let ustrythe other way. Ifyou are givena list ofnumbers, how can you saywhether it is
anA.P. or not?
For example, for anylist ofnumbers :
6, 9, 12, 15, . . . ,

Progressions 135
We check the difference ofthe succeeding terms. In the given list we have a2 – a1 = 9 – 6 = 3,
a3 – a2 = 12 – 9 = 3,
a4 – a3 = 15 – 12 = 3
We see that 2 1 3 2 4 3 ... 3
a a a a a a
- = - = - =
Here the difference ofanytwo consecutive terms in eachcase is 3. So, the given list is an
AP whose first terma is 6 and common difference d is 3.
For the list ofnumbers : 6, 3, 0, – 3, . . .,
a2 – a1 = 3 – 6 = – 3,
a3 – a2 = 0 – 3 = – 3
a4 – a3 = –3 – 0 = –3
2 1 3 2 4 3 3
a a a a a a
- = - = - =-
Similarly, this is also anAPwhose first termis 6 and the commondifference is –3.
So, weseethatifthedifferencebetweenanytwo consecutivetermsisconstant thenitisan
ArithmeticProgression.
In general, for anAP a1, a2, . . ., an, we can say
d = ak + 1 – ak where k ÎN; k > 1
where ak + 1 and ak are the (k + 1)th and the kth terms respectively.
Consider the list of numbers 1, 1, 2, 3, 5, . . . . By looking at it, you can tell that the
difference betweenanytwo consecutive terms is not the same. So, this is not anAP.
Note : To find d in theAP : 6, 3, 0, – 3, . . ., we have subtracted 6 from 3 and not 3 from 6.
We have to subtract the kth
termfromthe (k+ 1) thtermeven ifthe (k+ 1)th
termissmaller and
to find ‘d’ina givenAP, we need not findallof a2 - a1, a1 - a2 .... . It is enoughto find only one
ofthem
DO THIS
1. Take anyArithmetic Progression.
2. Add a fixed number to eachandevery termofAP. Writethe resulting numbers asa list.
3. Similarlysubtract a fixed number from eachand everytermofAP. Write the resulting
numbers as a list.
4. Multiplyor divide eachtermofAPbya fixednumber and write theresulting numbers as
a list.
5. Check whether the resulting lists areAPin each case.
6. What isyour conclusion?

Class-X Mathematics
136
Let us consider some examples
Example-1. For the AP :
1 1 3 5
, , ,
4 4 4 4
- - -
........, write the first term a and the common
difference d. And find the 7th
term
Solution : Here, a =
1
4
; d =
1 1 1
4 4 2
- -
- =
Remember that we can find d using any two consecutive terms, once we know that the
numbers are inAP.
The seventh termwould be
5 1 1 1 11
4 2 2 2 4
- -
- - - =
Example-2. Which ofthe following formsanAP? If theyformanAP, thenwrite thenext two
terms?
(i) 4, 10, 16, 22, . . . (ii) 1, – 1, – 3, – 5, . . . (iii) – 2, 2, – 2, 2, – 2, . . .
(iv) 1, 1, 1, 2, 2, 2, 3, 3, 3, . . . (v) x, 2x, 3x, 4x ......
Solution : (i) We have a2 – a1 = 10 – 4 = 6
a3 – a2 = 16 – 10 = 6
a4 – a3 = 22 – 16 = 6
i.e., ak + 1 – ak is same every time.
So, the given list ofnumbers forms anAPwiththe common difference d = 6.
The next two terms are: 22 + 6 = 28 and 28 + 6 = 34.
(ii) a2 – a1 = – 1 – 1 = – 2
a3 – a2 = – 3 – ( –1 ) = – 3 + 1 = – 2
a4 – a3 = – 5 – ( –3 ) = – 5 + 3 = – 2
i.e., ak + 1 – ak is same every time.
So, the given list ofnumbers forms anAPwiththe common difference d = – 2.
The next two terms are:
– 5 + (– 2 ) = – 7 and – 7 + (– 2 ) = – 9

Progressions 137
(iii) a2 – a1 = 2 – (– 2) = 2 + 2 = 4
a3 – a2 = – 2 – 2 = – 4
As a2 – a1 ¹ a3 – a2, the given list of numbers does not form anAP.
.
(iv) a2 – a1 = 1 – 1 = 0
a3 – a2 = 1 – 1 = 0
a4 – a3 = 2 – 1 = 1
Here, a2 – a1 = a3 – a2 ¹ a4 – a3.
So, the givenlist ofnumbers does not formanAP.
(v) We have a2 – a1 = 2x – x = x
a3 – a2 = 3x – 2x = x
a4 – a3 = 4x – 3x = x
i.e., ak+1 – ak is same every time.
So, the givenlist formanAP.
The next two terms are 4x + x = 5x and 5x + x = 6x.
EXERCISE - 6.1
1. Inwhichofthe following situations, doesthe list ofnumbersinvolvedformanarithmetic
progression, and why?
(i) The minimum taxi fare is ` 20 for the first km and there after ` 8 for each
additionalkm.
(ii) The amount ofair present in a cylinder whena vacuumpump removes
1
4
ofthe
air remaining in the cylinder at a time.
(iii) The cost ofdigging a well, after everymetre ofdigging, whenit costs `150 for
the first metre and rises by ` 50 for each subsequent metre.
(iv) The amount ofmoneyin the account everyyear, when `10000 is deposited at
compound interest at 8 % per annum.
2. Write first four terms oftheAP, when thefirst term a and the common difference d are
given as follows:
(i) a = 10, d = 10 (ii) a = –2, d = 0
(iii) a = 4, d = – 3 (iv) a = – 1, d =
1
2
(v) a = – 1.25, d = – 0.25

Class-X Mathematics
138
3. For thefollowingAPs, writethefirst termandthecommondifference:
(i) 3, 1, – 1, – 3, . . . (ii) – 5, – 1, 3, 7, . . .
(iii)
1 5 9 13
, , , ,....
3 3 3 3
(iv) 0.6, 1.7, 2.8, 3.9, . . .
4. Which of the following areAPs ? IftheyformanAP, find the common difference d and
write the next three terms.
(i) 2, 4, 8, 16, . . . (ii)
5 7
2, ,3, ,....
2 2
(iii) – 1.2, – 3.2, – 5.2, – 7.2, . . . (iv) – 10, – 6, – 2, 2, . . .
(v) 3, 3 2,3 2 2,3 3 2,....
+ + + (vi) 0.2, 0.22, 0.222, 0.2222, . . .
(vii) 0, – 4, – 8, –12, . . . (viii)
1 1 1 1
, , , ,....
2 2 2 2
- - - -
(ix) 1, 3, 9, 27, . . . (x) a, 2a, 3a, 4a, . . .
(xi) a, a2
, a3
, a4
, . . . (xii) 2, 8, 18, 32,.....
(xiii) 3, 6, 9, 12,.....
6.3 nth
TERM OF AN ARITHMETIC PROGRESSION
Let us consider the offer to Usha who applied for a job and got selected. She has been
offered a starting monthlysalaryof ` 8000, withanannualincrement of ` 500.What would be
her monthlysalaryofthe fifthyear?
To answer this,let us first see what her monthlysalary for the second year would be.
It would be ` (8000 + 500) = ` 8500.
In the same way, we can find the monthly salary for the 3rd
, 4th
and 5th
year by adding
` 500 to the salary ofthe previous year.
So, the salary for the 3rd
year = `(8500 + 500)
= ` (8000 + 500 + 500)
= ` (8000 + 2 × 500)
= ` [8000 + (3 – 1) × 500] (for the 3rd
year)
= ` 9000
Salary for the 4th
year = ` (9000 + 500)
= ` (8000 + 500 + 500 + 500)

Progressions 139
= ` (8000 + 3 × 500)
= ` [8000 + (4 – 1) × 500] (for the 4th
year)
= ` 9500
Salaryfor the 5th
year = ` (9500 + 500)
= ` (8000+500+500+500 + 500)
= ` (8000 + 4 × 500)
= ` [8000 + (5 – 1) × 500] (for the 5th
year)
= ` 10000
Observe that we are getting a list ofnumbers
8000, 8500, 9000, 9500, 10000, . . .
These numbers areinArithmetic Progression.
Looking at thepattern above, can we find her monthlysalaryin the 6th
year? The 15th
year?
And, assumingthat she is stillworking in the samejob, what wouldbe her monthlysalaryinthe
25th
year?Here we can calculate the salaryofthepresent year byadding ` 500 to the salaryof
previous year. Can we make this process shorter? Let us see. You mayhave alreadygot some
idea from theway we have obtained the salaries above.
Salary for the 15th
year = Salary for the 14th
year + ` 500
=
13 times
8000 500 500 500 ... 500 500
é ù
+ + + + + +
ê ú
ê ú
ë û
14444
24444
3
` `
= ` [8000 + 14 × 500]
= ` [8000 + (15 – 1) × 500] = ` 15000
i.e., First salary+ (15 – 1) ×Annual increment.
In the same way, her monthlysalaryfor the 25th
year would be
` [8000 + (25 – 1) × 500] = ` 20000
= First salary + (25 – 1) ×Annual increment
This example hasgivenusanidea about howto writethe 15th
term, orthe 25th
term.Byusing
the same idea, now let us find the nth
termof anAP.
Let a1, a2, a3, . . . be anAP whose first term a1 is a and the common difference is d.
Then,
the second term a2 = a + d = a + (2 – 1) d

Class-X Mathematics
140
the third term a3 = a2 + d = (a + d) + d = a + 2d = a + (3 – 1) d
the fourth term a4 = a3 + d = (a + 2d) + d = a + 3d = a + (4 – 1) d
. . . . . . . .
. . . . . . . .
Looking at the pattern, we can say that the nth
term an = a + (n – 1) d.
So, the nth
term of an AP with first term a and common difference d is given by
an = a + (n – 1) d.
an is also called the general term of theAP.
Ifthere are m terms intheAP, then am represents the last termwhich is sometimes also
denoted by l.
Finding terms of an AP: Using the above formula we can find different terms of an
arithemeticprogression.
Let usconsider some examples.
Example-3. Find the 10th
termof theAP : 5, 1, –3, –7 . . .
Solution : Here, a = 5, d = 1 – 5 = – 4 and n = 10.
We have an = a + (n – 1) d
So, a10 = 5 + (10 – 1) (–4) = 5 – 36 = –31
Therefore, the 10th
term of the givenAPis – 31.
Example-4. Which term of theAP : 21, 18, 15, . . . is – 81?
Is there anyterm0? Give reason for your answer.
Solution : Here, a = 21, d = 18 – 21 = – 3 and if an = – 81, we have to find n.
As an = a + ( n – 1) d,
we have – 81 = 21 + (n – 1)(– 3)
– 81 = 24 – 3n
– 105 = – 3n
So, n = 35
Therefore, the 35th
term of the givenAPis – 81.
Next, we want to know if there is anyn for which an = 0. If such n is there, then
21 + (n – 1) (–3) = 0,
i.e., 3(n – 1) = 21
i.e., n = 8
So, the eighthtermis 0.

Progressions 141
Example-5. Determine theAPwhose 3rd
termis 5and the 7th
termis 9.
Solution : We have
a3 = a + (3 – 1) d = a + 2d = 5 (1)
and a7 = a + (7 – 1) d = a + 6d = 9 (2)
Solving the pair oflinear equations (1) and(2), we get
a = 3, d = 1
Hence, the required AP is 3, 4, 5, 6, 7, . . .
Example-6. Check whether 301 is a term ofthe list of numbers 5, 11, 17, 23, . . .
Solution : We have :
a2 – a1 = 11 – 5 = 6, a3 – a2 = 17 – 11 = 6, a4 – a3 = 23 – 17 = 6
As (ak + 1 – ak) is the same for k = 1, 2, 3, etc., the given list of numbers is anAP.
Now, for thisAP we have a = 5 and d = 6.
We choose to beginwith the assumption that 301is nth termofthe thisAP.Wewillsee if
an ‘n’ exists for which an = 301.
We know
an = a + (n – 1) d
So, for 301 to be a termwe must have
301 = 5 + (n – 1) × 6
or 301 = 6n – 1
So, n =
302 151
6 3
=
But n should be apositive integer (Why?).
So, 301is not a termofthe given list of numbers.
Example-7. How manytwo-digit numbers are divisible by3?
Solution : The list oftwo-digit numbersdivisible by3 is :
12, 15, 18, . . . , 99
Is this anAP ? Yes it is. Here, a = 12, d = 3, an = 99.
As an = a + (n – 1) d,

Class-X Mathematics
142
we have 99 = 12 + (n – 1) × 3
i.e., 87 = (n – 1) × 3
i.e., n – 1 =
87
29
3
=
i.e., n = 29 + 1 = 30 (So, 99 is the 30th
term)
So, there are 30 two-digit numbers divisible by3.
termfromthe last ofthe theAPseriesgiven below :
AP : 10, 7, 4, . . ., – 62.
Solution : Here, a = 10, d = 7 – 10 = – 3, l = – 62,
where l = a + (n – 1) d
To find the 11th
termfrom the last term, we will find thetotal number of terms in theAP.
So, – 62 = 10 + (n – 1)(–3)
i.e., – 72 = (n – 1)(–3)
i.e., n – 1 = 24
or n = 25
So, there are 25 terms inthe givenAP.
The 11th
termfromthe last willbe the 15th
termofthe series. (Notethat it willnot be the
14th
term. Why?)
So, a15 = 10 + (15 – 1)(–3) = 10 – 42 = – 32
i.e., the 11th
termfrom the end is – 32.
Note : The 11th
termfromthe last is also equalto 11th
termoftheAPwithfirst term– 62and the
commondifference 3.
Example-9. Asumof ` 1000is invested at 8% simple interest per year. Calculate the interest
at the end ofeach year. Do these interests form anAP? If so, find the interest at the end of 30
years.
Solution : We knowthat the formula to calculate simple interest is given by
Simple Interest =
P R T
100
´ ´
So, the interest at the end of the 1st
year = `
1000 8 1
100
´ ´
= ` 80
The interest at the end of the 2nd
year = `
1000 8 2
100
´ ´
= ` 160

Progressions 143
The interest at the end ofthe 3rd
year =
1000 8 3
100
´ ´
= ` 240
Similarly, we canobtainthe interest at the end ofthe 4th
year, 5th
year, and so on. So,the
interest (in Rs) at the end of the 1st
, 2nd
, 3rd
, . . . years, respectivelyare
80, 160, 240, . . .
It is anAPas the difference between the consecutive terms inthe list is 80,
i.e., d = 80. Also, a = 80.
So, to find the interest at the end of30 years, we shallfind a30.
Now, a30 = a + (30 – 1) d = 80 + 29 × 80 = 2400
So, the interest at the end of 30 years willbe ` 2400.
Example-10. In a flower bed, there are 23 roseplants in the first row, 21 in the second, 19 in
the third, and so on. There are 5 rose plants in the last row. How many rows are there in the
flower bed?
Solution : The number of rose plants in the 1st
, 2nd
, 3rd
, . . ., rows are :
23, 21, 19, . . ., 5
It forms anAP (Why?).
Let the number ofrows in the flower bed be n.
Then a = 23, d = 21 – 23 = – 2, an = 5
As, an = a + (n – 1) d
We have, 5 = 23 + (n – 1)(– 2)
i.e., – 18 = (n – 1)(– 2)
i.e., n = 10
So, there are 10 rows in the flower bed.
EXERCISE - 6.2
1. Fill in the blanks in the following table, given that a is the first term, d the common
difference and an the nth termoftheAP:
S. No. a d n an
(i) 7 3 8 . . .
(ii) – 18 . . . 10 0

Class-X Mathematics
144
(iii) . . . – 3 18 – 5
(iv) – 18.9 2.5 . . . 3.6
(v) 3.5 0 105 . . .
2. Find the
(i) 30th
term of theAP 10, 7, 4 ......
(ii) 11th
termof theAP :
1
3, , 2,.....
2
-
-
3. Find therespective terms for the followingAPs.
(i) a1
= 2, a3
= 26 find a2
(ii) a2
= 13, a4
= 3 find a1
, a3
(iii) a1
= 5, a4
= 9
1
2
find a2
, a3
(iv) a1
= -4, a6
= 6 find a2
, a3
, a4
, a5
(v) a2
= 38, a6
= -22 find a1
, a3
, a4
, a5
4. Which term of the AP : 3, 8, 13, 18, . . . ,is 78?
5. Find the number ofterms ineachofthe followingAPs :
(i) 7, 13, 19, . . . , 205 (ii) 18, 15
1
2
, 13, ..., -47
6. Check whether, –150 is a term of theAP : 11, 8, 5, 2 . . .
7. Find the 31st
term ofanAP whose 11th
term is 38 and the 16th
termis 73.
8. Ifthe 3rd
and the 9th
terms ofanAPare 4 and – 8 respectively, which termofthisAPis
zero?
9. The 17th
term of anAPexceeds its 10th
termby 7. Find the common difference.
10. TwoAPshavethesamecommondifference.Thedifferencebetween their 100th
termsis
100. What is the difference between their 1000th
terms?
11. How manythree-digit numbers are divisibleby7?
12. How manymultiples of4 lie between10 and 250?
13. For what value of n, are the nth
terms of two APs: 63, 65, 67, . .. and 3, 10, 17, . . .
equal?
14. Determine theAPwhose third termis 16 and the 7th
termexceeds the 5th
termby 12.
15. Find the 20th
term fromthe end of theAP: 3, 8, 13, . . ., 253.

Progressions 145
16. The sumof the 4th
and 8th
terms ofanAP is 24 and the sum of the 6th
and 10th
terms is
44. Find the first three terms of theAP.
17. Subba Rao started his job in 1995 at a monthly salary of ` 5000 and received an
increment of ` 200 each year. In which year did his salaryreach ` 7000?
6.4 SUM OF FIRST n TERMS IN ARITHMETIC PROGRESSION
Let us consider the situation again given in Section 6.1 in
which Hemaput ` 1000 moneyboxwhenher daughter was one
year old, ` 1500 on her second birthday, ` 2000 on her third
birthdayandwillcontinue inthesameway.Howmuchmoneywill
be collected inthe moneybox bythe time her daughter is 21 years
old?
Here, the amount ofmoney (in Rupees) put in the money
boxonherfirst,second,third,fourth...birthdaywere respectively
1000, 1500, 2000, 2500, . . . till her 21st
birthday. To find the
totalamount inthe moneyboxon her 21st
birthday, we willhave to write each ofthe 21 numbers
in thelist above and thenadd themup. Don’t you think it would bea tedious and timeconsuming
process? Can we make the process shorter?
This would be possible ifwe can find a method for getting this sum. Let us see.
6.4.1 HOW ‘GAUSS’ FOUND THE SUM OF TERMS
We consider the problem given to Gauss, to solve when he was
just 10 years old. He was asked to find the sum ofthe positive integers
from1 to 100. He repliedthat the sumis5050. Canyou guesshow could
he do it?
Let S = 1 + 2 + 3 + . . . + 99 + 100
And then, reverse the numbers to write
S = 100 + 99 + . . . + 3 + 2 + 1
When he added these two, term bytermhe got,
2S = (100 + 1) + (99 + 2) + . . . + (3 + 98) + (2 + 99) + (1 + 100)
= 101 + 101 + . . . + 101 + 101 (100 times) (check this out and discuss)
So, S =
100 101
2
´
= 5050, i.e., the sum = 5050.
Carl Fredrich Gauss
(1777-1855) was a great
GermanMathematician

Class-X Mathematics
146
6.4.2 SUM OF n TERMS OF AN AP.
We willnow use the sametechnique that was used byGauss to find the sumofthe first n
terms ofanAP:
a, a + d, a + 2d, . . .
The nth
term of this AP is a + (n – 1) d.
Let Sn denote the sum of the first n terms of the A.P.
Bywriting Sn in two different orders, we get
n
S ( ) ( 2 ) ... ( 1)
a a d a d a n d
= + + + + + + + -
n
S [ ( 1) ] [ ( 2) ] ...
a n d a n d a
= + - + + - + +
Adding termbyterm
2Sn = [2a + (n - 1)d ] + [2a + (n - 1)d ] + .... + [2a + (n - 1)d ] n times
= n[2a + (n - 1)d ]
Sn [2 ( 1) ]
2
n
a n d
= + - [ { ( 1) }]
2
n
a a n d
= + + -
2
n
= [first term+ nth
term] ( )
2
= + n
n
a a
If the first and last terms ofanAP are given andthe common difference is not given then
Sn ( )
2
= + n
n
a a is very usefulto find Sn or Sn ( )
2
n
a l
= + where ‘l’ is the last term.
Now wereturnto the example (c) inthe introduction6.1. The amount ofmoney(inRs)
in the money box of Hema’s daughter on 1st
, 2nd
, 3rd
, 4th
birthday, . . ., were 1000, 1500,
2000, 2500, . . ., respectively.
This is anAP.We have to find the totalmoneycollected on her 21st
birthday, i.e., the sum
ofthe first 21 terms ofthisAP.
Here, a = 1000, d = 500and n = 21. Using the formula :
Sn
[2 ( 1) ],
2
n
a n d
= + -
we have S =
21
[2 1000 (21 1) 500]
2
´ + - ´
=
21
[2000 10000]
2
+

Progressions 147
=
21
[12000] 126000
2
=
So, the amount of moneycollected on her 21st
birthdayis ` 1,26,000.
We use Sn in place ofS to denote the sumoffirst n terms oftheAPso that we know how
manyterms we have added. We write S20 to denote the sumofthe first 20 terms ofan AP. The
formula for the sumofthe first n terms involves four quantities Sn, a, d and n. Ifwe know any
three ofthem, we can findthe fourth.
Remark : The nth
termofanAPisthe difference ofthe sumto first n termsand the sumto first
(n – 1) terms of it, i.e., an = Sn – Sn – 1.
DO THIS
Find the sumof indicated number ofterms in each ofthe followingAPs
(i) 16, 11, 6 .....; 23 terms (ii) -0.5, -1.0, -1.5, .....; 10 terms
(iii)
1 3
1, , .....,
4 2
- 10 terms
Example-11. If the sumof the first 14 terms of anAP is 1050 and its first term is 10, find the
20th
term.
Solution : Here, Sn = 1050; n = 14, a = 10
Sn = [2 ( 1) ]
2
n
a n d
+ -
1050 =
14
[2 13 ] 140 91
2
a d d
+ = +
910 = 91d
d = 10
a20 = 10 + (20 -1) 10 = 200
Example-12. How manytermsoftheAP: 24, 21, 18, . . . must be taken so that their sumis 78?
Solution : Here, a = 24, d = 21 – 24 = –3, Sn = 78. Let number of terms ofAP is n, then we
need to find n.
We know that Sn = [2 ( 1) ]
2
n
a n d
+ -
So, 78 [48 ( 1)( 3)] [51 3 ]
2 2
n n
n n
= + - - = -
or 3n2
– 51n + 156 = 0

Class-X Mathematics
148
or n2
– 17n + 52 = 0
or (n – 4)(n – 13) = 0
or n = 4 or 13
Both values of n are admissible. So, the number ofterms is either 4or 13.
Remarks :
1. In this case, the sum ofthe first 4 terms = the sumofthe first 13 terms = 78.
2. Two answers are possible because the sum of the terms from 5th
to 13th
willbe zero.
This is because a is positive and d is negative, so that some termsare positiveand some
are negative, and willcancelout eachother.
Example-13. Find the sumof:
(i) the first 1000 naturalnumbers (ii) the first n naturalnumbers
Solution :
(i) Let S = 1 + 2 + 3 + . . . + 1000
Using the formula Sn = ( )
2
n
a l
+ for the sumofthe first n terms of anAP, we have
S1000 =
1000
(1 1000)
2
+ = 500 × 1001 = 500500
So, the sumofthe first 1000 positive integers is 500500.
(ii) Let Sn = 1 + 2 + 3 + . . . + n
Here a = 1 and the last term l is n.
Therefore, Sn =
(1 )
2
n n
+
(or) Sn =
( 1)
2
n n +
So, the sum of first n positive integers is given by
Sn =
( 1)
2
n n +
Example-14. Find the sumoffirst 24 terms ofthe list ofnumbers whose nth
term is given by
an = 3 + 2n
Solution : As an = 3 + 2n,
so, a1 = 3 + 2 = 5

Progressions 149
a2 = 3 + 2 × 2 = 7
a3 = 3 + 2 × 3 = 9
. . .
List of numbers becomes 5, 7, 9, 11, . . .
Here, 7 – 5 = 9 – 7 = 11 – 9 = 2 and so on.
So, it forms anAPwith common difference d = 2.
To find S24, we have n = 24, a = 5, d = 2.
Therefore, S24 =
24
[2 5 (24 1) 2] 12(10 46) 672
2
´ + - ´ = + =
So, sumoffirst 24 terms ofthe list ofnumbers is 672.
Example-15. Amanufacturer ofTVsets produced 600 sets in the third year and 700 setsinthe
seventh year. Assuming that the production increases uniformlybya fixed number every year,
find:
(i) the production in the 1st
year (ii) the production in the 10th
year
(iii) the totalproductionin first 7 years
Solution : (i) Since theproduction increases uniformlybya fixed number everyyear, thenumber
ofTV sets manufactured in 1st
, 2nd
, 3rd
, . . ., years willform anAP.
Let usdenote the number ofTV sets manufactured inthe nth year byan.
Then, a3 = 600 and a7 = 700
or, a + 2d = 600
and a + 6d = 700
Solving these equations, we get d = 25 and a = 550.
Therefore, productionofTV sets in the first year is 550.
(ii) Now a10 = a + 9d = 550 + 9 × 25 = 775
So, production of TV sets inthe 10th
year is 775.
(iii) Also, S7 =
7
[2 550 (7 1) 25]
2
´ + - ´
7
[1100 150] 4375
2
= + =
Thus, the totalproduction ofTVsets in first 7 years is 4375.

Class-X Mathematics
150
EXERCISE - 6.3
1. Find the sumofthe followingAPs:
(i) 2, 7, 12, . . ., to 10 terms. (ii) –37, –33,–29, .. .,to 12terms.
(iii) 0.6, 1.7, 2.8, . . ., to 100 terms. (iv)
1 1 1
, , , .....,to 11terms.
15 12 10
2. Find the sumsgiven below :
(i)
1
7 10 14 .... 84
2
+ + + + (ii) 34 + 32 + 30 + . . . + 10
(iii) –5 + (–8) + (–11) + . . . + (–230)
3. In anAP:
(i) given a = 5, d = 3, an
= 50, find n and Sn
.
(ii) given a = 7, a13
= 35, find d and S13
.
(iii) given a12
= 37, d = 3, find a and S12
.
(iv) given a3
= 15, S10
= 125, find d and a10
.
(v) given a = 2, d = 8, Sn
= 90, find n and an
.
(vi) given an
= 4, d = 2, Sn
= –14, find n and a.
(vii) given l = 28, S = 144, and there are total 9 terms, find a.
4. Thefirst andthelasttermsofanAPare17and350respectively. Ifthecommondifference
is 9, how manyterms are there and what is their sum?
5. Find the sum of first 51 terms of anAP whose second and third terms are 14 and 18
respectively.
6. If thesum of first 7 terms of anAPis 49 and that of 17 terms is289, find the sumof first
n terms.
7. Show that a1, a2, . . ., an, . . . form anAP where an is defined as below :
(i) an = 3 + 4n (ii) an = 9 – 5n
Also find the sumofthefirst 15 terms ineach case.
8. Ifthe sumofthe first n termsofanAPis4n – n2
, what is the first term(notethat the first
termis S1)? What isthe sumoffirst two terms?What isthe secondterm?Similarly, find
the 3rd, the 10th and the nth terms.
9. Find the sumofthe first 40 positive integers divisibleby6.
10. A sumof ` 700 is to be used to give seven cash prizes to students ofa schoolfor their
overallacademic performance. Ifeachprize is ` 20less thanits preceding prize,findthe
value ofeachofthe prizes.

Progressions 151
11. In a school, students thought ofplanting trees in and around the school to reduce air
pollution. It was decided that the number of trees, that each section of each class will
plant, willbe the same as the class, in which theyarestudying, e.g., a sectionofClass I
willplant 1 tree, a sectionofClass II willplant 2 treesandso ontillClassXII. There are
three sections ofeach class. How manytrees willbe planted by the students?
12. A spiral is made up of successive
semicircles, with centres alternatelyat A
and B, starting with centre at A, of radii
0.5 cm, 1.0 cm, 1.5 cm, 2.0 cm, . . . as
shown inFigure. What isthe totallength
of such a spiral made up of thirteen
consecutive semicircles?(Take
22
7
p = )
[Hint : Length ofsuccessive semicircles is
l1, l2, l3, l4, . . . with centres atA, B,A, B, . . ., respectively.]
13. 200 logs arestacked in the following manner: 20 logs inthe bottomrow, 19 in the next
row, 18 in the row next to it and so on. In how many rows are the 200 logsplaced and
how manylogs are in the top row?
14. In abucket and ballrace, a bucket is placed at the starting point, which is 5 mfromthe
first ball, and the other balls are placed 3 mapart in a straight line. There are tenballs in
the line.
Acompetitor starts from the bucket, picks up the nearest ball, runs back withit, drops it
in the bucket, runsback to pick up the next ball, runsto thebucket to drop it in, and she
continues inthe same way untilallthe ballsare inthe bucket. What is the total distance
the competitor has to run?
[Hint : To pick upthe first balland thesecond ball, the totaldistance(in metres) run by
a competitor is 2 × 5 + 2 × (5 + 3)]
5m 3m 3m
A B
l1
l3
l2
l4

Class-X Mathematics
152
6.5 GEOMETRIC PROGRESSIONS
Consider thelists
(i) 30, 90, 270, 810 ..... (ii)
1 1 1 1
, , , .....
4 16 64 256
(iii) 30, 24, 19.2, 15.36, 12.288
Can we write the next term in each of the lists above?
In (i), eachtermis obtained bymultiplying the preceeding termby3.
In (ii), each termisobtained bymultiplying the preceeding term by
1
4
.
In (iii), each term is obtained bymultiplying the preceedingterm by0.8.
Inallthe lists given above, we see that successive terms are obtained bymultiplying the
preceedingtermbyafixed number. Suchalist ofnumbersis saidto formGeometricProgression
(GP).
This fixed number is called the commonration‘r’ofGP. So in the above example (i), (ii),
(iii) the common ratios are 3,
1
4
, 0.8 respectively.
.
Let us denote the first term of a GP by a and common ratio r. To get the second term
according to therule ofGeometricProgression, wehave tomultiplythefirst termbythecommon
ratio r, where a ¹ 0, r ¹ 0 and r ¹ 1
The second term= ar
Third term = ar. r = ar2
a, ar, ar2
..... is called the general form ofa GP.
In the aboveGPthe ratio betweenanyterm(except first term) and its preceding termis ‘r’
i.e.,
2
..........
ar ar
r
a ar
= = =
If we denote the first term ofGP by a1, second termby a2 ..... nth
termby an
then 2 3
1 2 1
......
-
= = = =
n
n
a a a
r
a a a
Alist of numbers a1, a2, a3 .... an ... is called a geometric progression (GP), if each
termis nonzero and
1
n
n
a
r
a -
= (r ¹ 1)
where n is a natural number and n > 2.

Progressions 153
DO THIS
Find whichofthe following are not GPs
1. 6, 12, 24, 48, ..... 2. 1, 4, 9, 16, ......
3. 1, -1, 1, -1, ..... 4. -4, -20, -100, -500, .....
Some more example of GP are :
(i) Apersonwritesaletter tofourofhisfriends. He askseachone ofthemtocopythe letter
and give it to four different persons with same instructions so that they can move the
chainaheadsimilarly. Assumingthat thechainisnot brokenthenumber ofletters at first,
second, third... stages are
1, 4, 16, 64, 256 .............. respectively.
(ii) The totalamount at the end offirst , second, third .... year if ` 500/- is deposited in the
bank withannualrate 10% interest compounded annuallyis
550, 605, 665.5 ......
(iii) Asquareisdrawnbyjoiningthe mid pointsofthe sides
of a given square. Athird square is drawn inside the
second square in the same way and this process
continues indefinitely. If a side of the first square is
16cm then the area of first, second, third ..... square
willbe respectively.
256, 128, 64, 32, .....
(iv) Initiallya pendulumswings through anarc of 18 cms.
On each successive swing the length of the arc is 0.9th
of the
previouslength. So thelengthofthearcatfirst, second, third.......
swing willberesepectively(incm).
18, 16.2, 14.58, 13.122......
THINK - DISCUSS
1. Explainwhyeachofthe lists aboveis a GP.
2. To knowabout a GP, what is the minimuminformationthat we need?

Class-X Mathematics
154
Now let uslearnhow to construct a GP. whenthe first term‘a’andcommonratio ‘r’are
given. And also learn how to check whether thegiven list ofnumbers is a GP.
Example-16. Write the GP. ifthe first terma = 3, and the common ratio r = 2.
Solution : Since ‘a’ isthe first termit caneasily be written
We know that in GP. every succeeding term is obtained bymultiplying the preceding
termwith common ratio ‘r’. So to get the second termwe have to multiplythe first term a = 3
bythe common ratio r = 2.
Second term = ar = 3 × 2 = 6 (Q First term× commonratio)
Similarlythethirdterm= second term× common ratio
= 6 × 2 = 12
Ifwe proceedinthis waywe get the following G.P.
3, 6, 12, 24,..... .
Example-17. Write GP. if a = 256, r =
1
2
-
Solution : General form of GP = a, ar, ar2
, ar3
, .....
= 256, 256
1
2
-
æ ö
ç ÷
è ø
, 256
2
1
2
-
æ ö
ç ÷
è ø
, 256
3
1
2
-
æ ö
ç ÷
è ø
= 256, -128, 64, -32 ......
Example-18. Find the common ratio ofthe GP 25, -5, 1,
1
5
-
.
Solution : We know that ifthe first, second, third .... terms ofa GPare a1, a2, a3 .... respectively
the common ratio 3
2
1 2
.....
a
a
r
a a
= = =
Here a1 = 25, a2 = -5, a3 = 1.
So commonratio
5 1 1
25 5 5
r
- -
= = =
-
.
Example-19. Whichofthe following lists ofnumbers formGP?
(i) 3, 6, 12, ..... (ii) 64, -32, 16,
(iii)
1 1 1
, , ,......
64 32 8

Progressions 155
Solution : (i)We know that a list ofnumbers a1, a2, a3, .....an ..... is called a GP ifeach termis
non zero and
3
2
1 2 1
..... n
n
a a
a
r
a a a -
= = =
Here all the terms are non zero. Further
2
1
6
2
3
a
a
= = and
3
2
12
2
6
a
a
= =
i.e.,
3
2
1 2
2
a
a
a a
= =
So, the givenlist ofnumber forms a GP with the common ratio 2.
(ii) Allthe terms are non zero.
2
1
32 1
64 2
a
a
- -
= =
and
3
1
16 1
32 2
a
a
-
= =
-
3
2
1 2
1
2
a
a
a a
-
= =
So, thegivenlist ofnumbersforma GP withcommonratio
1
2
-
.
(iii) Allthe terms are non zero.
2
1
1
32 2
1
64
a
a
= =
3
2
1
8 4
1
32
a
a
= =
Here
3
2
1 2
a
a
a a
¹
So, the givenlist ofnumbers doesnot formGP.

Class-X Mathematics
156
EXERCISE - 6.4
1. Inwhichofthefollowing situations, doesthe list ofnumbersinvolvedisinthe formofaGP?
(i) SalaryofSharmila, whenhersalaryis `5,00,000for the first year and expected
to receive yearlyincrease of 10% .
(ii) Number ofbricks needed to make each step, ifthe stair case hastotal30 steps,
provided that bottom step needs 100 bricks and each successive step needs 2
brick less thanthe previous step.
(iii) Perimeter of the each
triangle, when the mid
points of sides of an
equilateral triangle
whosesideis24 cmare
joined to form another
triangle, whose mid
pointsinturnarejoined
to form still another
triangleandtheprocess
continuesindefinitely.
2. Write three terms ofthe GPwhen the first term‘a’and the common ratio ‘r’are given?
(i) a = 4; r = 3 (ii) 5
a = ;
1
5
r =
(iii) a = 81; r =
1
3
-
(iv)
1
64
a = ; r = 2
3. Which ofthe following are GP?Iftheyare inGP. Write next three terms?
(i) 4, 8, 16 ..... (ii)
1 1 1
, , .....
3 6 12
-
(iii) 5, 55, 555, .... (iv) -2, -6, -18 .....
(v)
1 1 1
, , .....
2 4 6
(vi) 3, -32
, 33
, .....
(vii)
1
, 1, ,....., ( 0)
x x
x
¹ (viii)
1
, 2, 4 2.....
2
-
(ix) 0.4, 0.04, 0.004, .....
4. Find x so that x, x + 2, x + 6 are consecutive terms ofa geometric progression.
24 cm 24 cm
24 cm

Progressions 157
6.6 nth
TERM OF A GP
Let us examinea problem. The number ofbacteria ina certain culturetriples everyhour.
Iftherewere 30 bacteriapresentinthecultureoriginally,then, what wouldbenumber ofbacteria
infourthhour?
To answer this let us first see what the number ofbacteria in second hour would be.
Since for everyhour it triples
No.of bacteria in Second hour = 3 × no.of bacteria infirst hour
= 3 × 30 = 30 × 31
= 30 × 3(2-1)
= 90
No.ofbacteriainthird hour = 3 × no.of bacteria in second hour
= 3 × 90 = 30 × (3×3)
= 30 × 32
= 30 × 3(3-1)
= 270
No.ofbacteriain fourth hour = 3 × no.ofbacteriain third hour
= 3 × 270 = 30 × (3×3×3)
= 30 × 33
= 30 × 3(4-1)
= 810
Observe that we are getting a list of numbers
30, 90, 270, 810, ....
These numbers are in GP (why?)
Now looking at thepatternformed above, can you find number of bacteria in20th
hour?
You may have already got some idea from the way we have obtained the number of
bacteria as above. Byusing the same pattern, we can compute that number ofbacteria in 20th
hour.
19 terms
30 (3 3 ... 3)
= ´ ´ ´ ´
14
4
244
3
= 30 × 319
= 30 × 3(20-1)
This example would have given you some idea about how to write the 25th
term, 35th
termand more generallythe nthtermofthe GP.
Let a1, a2, a3 ..... be in GP whose ‘first term’a1 is a and the commonratio is r.

Class-X Mathematics
158
then the second term a2 = ar = ar(2-1)
thethirdterm a3 = a2×r = (ar) × r = ar2
= ar(3-1)
the fourthterm a4 = a3×r = ar2
× r = ar3
= ar(4-1)
................................................
................................................
Looking at the patternwe cansaythat nth
term an = arn-1
So nth
termof a GP with first term ‘a’and common ratio ‘r’is given by an = arn-1
.
Let us consider some examples
and nth
termofthe GP.
5 5 5
, , .......
2 4 8
Solution : Here a =
5
2
and
5
1
4
5 2
2
r = =
Then
19
20 1
20 20
5 1 5
2 2 2
a ar - æ ö
= = =
ç ÷
è ø
and
1
1 5 1 5
2 2 2
n
n
n n
a ar
-
- æ ö
= = =
ç ÷
è ø
Example-21. Which term of the GP : 2, 2 2 , 4 ..... is 128 ?
Solution : Here a = 2 r =
2 2
2
2
=
Let 128 be the nth
termofthe GP.
Then an = arn-1
= 128
1
2.( 2) 128
n-
=
1
( 2) 64
n-
=
1
6
2
(2) 2
n-
=

Progressions 159
Þ
1
6
2
n -
=
n = 13.
Hence 128 is the 13th
termofthe GP.
Example-22. In a GP the 3rd
termis 24 and 6th
term is 192. Find the 10th
term.
Solution : Here a3 = ar2
= 24 ...(1)
a6 = ar5
= 192 ...(2)
Dividing (2) by(1) we get
5
2
192
24
ar
ar
=
Þ r3
= 8 = 23
Þ r = 2
Substituting r = 2 in (1) we get a = 6.
a10 = ar9
= 6(2)9
= 3072.
EXERCISE-6.5
1. For each geometric progressionfind the common ratio ‘r’, and thenfind an
(i)
3 3 3
3, , , .........
2 4 8
(ii) 2, -6, 18, -54
(iii) -1, -3, -9, -27 .... (iv)
4 8
5, 2, , .........
5 25
2. Find the 10th
and nth
term of GP. : 5, 25, 125, .....
3. Find the indicated term of eachGeometric Progression
(i) a1 = 9;
1
;
3
r = find a7 (ii) a1 = -12;
1
;
3
r = find a6
4. Whichtermofthe GP.
(i) 2, 8, 32, ..... is 512 ? (ii) 3, 3, 3 3 ...... is 729 ?
(iii)
1 1 1
, , .....
3 9 27
is
1
2187
?

Class-X Mathematics
160
5. Find the 12th
term ofa GP. whose 8th
term is 192 and the common ratio is 2.
6. The 4th
term of a geometric progression is
2
3
and the seventh term is
16
81
. Find the
geometric series.
7. If the geometric progressions 162, 54, 18 ..... and
2 2 2
, , ....
81 27 9
have their nth
term
equal. Find the value of n.
OPTIONAL EXERCISE
1. Which termoftheAP: 121, 117, 113, . . ., is the
first negative term?
[Hint : Find n for an < 0]
2. The sumofthe thirdand the seventhterms ofan
AP is 6 and their product is 8. Find the sumof
first sixteentermsoftheAP.
3. A ladder has rungs 25 cm apart. The rungs
decrease uniformlyinlength from45 cmat the
bottom to 25 cm at the top. If the top and the
bottomrungs are
1
2
2
mapart,what is thelength
ofthe wood required for the rungs?
[Hint : Number ofrungs=
250
1
25
+ ]
4. The housesofa row arenumbered consecutively
from1 to 49. Show that there is a value ofx such
thatthesumofthenumbersofthehousespreceding
the house numbered x is equalto the sumofthe numbers ofthe housesfollowing it.And
find this value of x.
[Hint : Sx – 1 = S49 – Sx]
5. A smallterrace at a footballground comprises of15 steps eachof which is 50mlong and
built ofsolid bricks.
25 cm
45 cm
1
2
2
m
25 cm

Progressions 161
Eachstephasa rise of
1
4
mand atreadof
1
2
m. (seeFig. 5.8). Calculate the totalvolume
of the terrace.
[Hint : Volume ofthe first step =
1 1
50
4 2
´ ´ m3
]
6. 150 workers were engaged to finish a piece ofwork in a certain number ofdays. Four
workers dropped fromthe work on the second day. Four workers dropped on third
dayandso on. It took 8 more daysto finishthe work.Find the number ofdaysinwhich
the work was completed.
[Let the no.ofdaysto finish the work is ‘x’ then
8
150 [2 150 ( 8 1)( 4)]
2
x
x x
+
= ´ + + - - ]
[Ans. x = 17 Þ x + 8 = 17 + 8 = 25]
7. Amachine costs ` 5,00,000. If its value depreciates 15% in thefirst year, 13
1
2
%in the
second year, 12% in the third year and so on. What will be its value at the end of 10
years, whenallthe percentages willbe applied to the originalcost?
[Total depreciation = 15 + 13
1
2
+12+....10terms.
10
S [30 13.5] 82.5%
2
n = - = ]
after 10 year original cost =100 - 82.5 = 17.5 i.e., 17.5% of 5,00,000
1
m
4
1
m
2
5
0
m

Class-X Mathematics
162
In this chapter, you have studied the following points :
1. An arithmetic progression (AP) isa list ofnumbersinwhicheach termis obtained by
adding a fixed number d to thepreceding term, except thefirst term. The fixednumber d
is called the common difference.
The terms ofAP are a, a + d, a + 2d, a + 3d, . . .
2. Agiven list ofnumbers a1, a2, a3, . . . is anAP, ifthe differences a2 – a1, a3 – a2, a4 – a3,
. . ., give the same value, i.e., if ak + 1 – ak is the same for different values of k.
3. In anAPwith first terma and commondifference d, the nth
term(orthe generalterm) is
given by an = a + (n – 1) d.
4. The sumofthe first n terms ofanAPis givenby:
S [2 ( 1) ]
2
n
a n d
= + -
5. Ifl is the last termofthe finiteAP, saythe nth
term, thenthe sumofalltermsoftheAPis
givenby:
S ( )
2
n
a l
= + .
6. A Geometric Progression (GP) isa list ofnumbersinwhichfirst termis non-zero each
succeeding termisobtained bymultiplying preceeding termwith afixednonzero number
‘r’except first term. This fixed number is called common ratio ‘r’.
The general formof GP is a, ar, ar2
, ar3
....
7. If the first term and common ratio of a GP are a and r respectively then nth term
an = arn-1
.
Suggested Projects
1. Write downa sequence and verifywhether it isAPor not, using a grid paper.
2. Find the sum ofnterms in anAPusing grid paper.

7.1 INTRODUCTION
You know that in chess, the Knight moves in ‘L’shape or two and a halfsteps (see figure).
It can jump over other pieces too. A Bishop moves
diagonally, as manysteps as are free infront of it.
Find out how other pieces move. Also locate
Knight, Bishop and other pieces on the board and see
how theymove.
Consider that the Knight is at the origin (0, 0). It can
movein4directionsasshownbydottedlinesinthefigure.
Findthecoordinatesofitspositionafterthevariousmoves
showninthe figure.
DO THIS
i. Fromthe figure write coordinates ofthe pointsA, B, C, D, E, F, G, H.
ii. Find the distance covered bythe Knight in eachofits8movesi.e. find thedistance ofA,
B, C, D, E, F, G and H fromthe origin.
iii. What is the distance between two points H and C? and also find the distance between
two pointsAand B
7.2 DISTANCE BETWEEN TWO POINTS
The two points (2, 0) and (6, 0) lie on the X-axis as shownin figure.
It is easy to see that the distance between two pointsAand B as 4 units.
We can say the distance between points lying on X-axis is the difference between the
x-coordinates.
Coordinate Geometry
7
A B
C
D
E
F
G
H
a b c d e f g h
8
7
6
5
4
3
2
1

Class-X Mathematics
164
What is the distance between
(-2, 0) and (-6, 0)?
The difference in the value of
x-coordinates is
(-6) - (-2) = -4 (Negative)
We never say the distance in
negative values.
So, we willtake absolute value of
the difference.
Therefore, the distance
= | (- 6) - (-2)| = |-4| = 4 units.
So, in general for the points
A(x1, 0), B(x2, 0) on the X-axis, the
distance betweenAand B is |x2 - x1|
Similarly, if two points lie on
Y-axis, then the distance between the
pointsAand B would be the difference
betweentheir ycoordinatesofthepoints.
The distancebetween two points
(0, y1) (0, y2) would be |y2 - y1|.
For example, let the points
beA(0, 2) and B(0, 7)
Then, the distance betweenAand
B is |7 - 2| = 5 units.
DO THIS
1. Where do these following points lie (-4, 0), (2, 0), (6, 0) and (-8, 0) on coordinate
plane?
2. What is the distance between points (-4, 0) and (6, 0) ?
-9
Y
O
X
1
Y
1
X
-8 -7 -6 -5 -4 -3 -2 -1 1 2 3 4 5 6 7 8 9
-1
-2
-3
-4
-5
-6
-7
-8
-9
1
2
3
4
5
6
7
8
9
A(2, 0) B(6, 0)
-9
Y
O
X
1
Y
1
X
-8 -7 -6 -5 -4 -3 -2 -1 1 2 3 4 5 6 7 8 9
-1
-2
-3
-4
-5
-6
-7
-8
-9
1
2
3
4
5
6
7
8
9
A (0, 2)
B (0, 7)
Scale
X-axis : 1 cm = 1 unit
Y-axis : 1 cm = 1 unit
Scale

Coordinate Geometry 165
TRY THIS
1. Where do these following points lie (0, -3), (0, -8), (0, 6) and (0, 4) on coordinate
plane?
2. What is the distance between(0, -3), (0, -8) and justifythat the distancebetweentwo
points onY-axis is |y2 - y1| oncoordinate plane?
THINK - DISCUSS
How willyou find the distance betweentwo points inwhichxor ycoordinatesare same
but not zero?
7.3 DISTANCE BETWEEN TWO POINTS ON A LINE PARALLEL TO THE
COORDINATE AXES.
Consider the pointsA(x1, y1) and B(x2, y1). Since the y-coordinates areequal, points lie
on a line, parallelto X-axis.
AP and BQ are drawn
perpendicular to X-axis.
Observe the figure. APQB is a
rectangle.
Therefore, AB = PQ.
PQ = |x2 - x1| (i.e., The modulus
of difference betweenxcoordinates)
Similarly, line joining two points
A(x1, y1) and B(x1, y2) is parallel to
Y-axis.Thenthedistancebetweenthese
two points is |y2 - y1| (It is read as
modulus of the difference of y
coordinates).
Y
X
X¢
Y¢
A( , )
x y
1 1 B( , )
x y
2 1
( - )
x x
2 1
O P Q
( - )
x x
2 1

Class-X Mathematics
166
Example-1. What is the distance between A(4,0) and B (8, 0).
Solution : The abosolutevalue ofthe differenceinthe x coordinatesis|x2 -x1| =|8 -4|=4 units.
Example-2. Aand B are two points given by (8, 3), (-4, 3). Find the distance between
A and B.
Solution : Here x1 and x2 are lying intwo different quadrants and y-coordinate are equal.
Distance AB = |x2 - x1| = |-4 - 8| = |-12| = 12 units
DO THIS
Find the distance betweenthe following points.
i. (3, 8), (6, 8) ii. (-4,-3), (-8,-3) iii. (3, 4), (3, 8) (iv) (-5, -8), (-5, -12)
Let A and B denote the points
(4, 0) and (0, 3) and ‘O’ be the
origin.
The DAOBis aright angle triangle.
Fromthe figure
OA = 4 units (x-coordinate)
OB = 3 units (y-coordinate)
Then distanceAB = ?
Hence, by using Pythagorean
theorem
AB2
= AO2
+ OB2
AB2
= 42
+ 32
AB = 16 9 25 5units
+ = =
is the distance betweenAand B.
DO THIS
Find the distancebetween the following points (i)A= (2, 0) and B(0, 4) (ii) P(0, 5) and
Q(12, 0)
TRY THIS
Find the distance between points ‘O’(origin) and ‘A’(7, 4).
-9
Y
O
X
1
Y
1
X
-8 -7 -6 -5 -4 -3 -2 -1 1 2 3 4 5 6 7 8 9
-1
-2
-3
-4
-5
-6
-7
-8
-9
1
2
3
4
5
6
7
8
9
A (4, 0)
B (0, 3)
Scale

THINK - DISCUSS
1. Ramu says the distance of a point P(x1, y1) from the origin O(0, 0) is 2 2
1 1
x y
+ . Do
you agree withRamu or not? Why?
2. Ramu also writes the distance formulasAB = 2 2
1 2 1 2
( ) ( )
x x y y
- + - (why?)
7.4 DISTANCE BETWEEN ANY TWO POINTS IN THE X-Y PLANE
Let A(x1, y1) and B(x2, y2) be anytwo points ina plane as shown in figure.
DrawAPand BQ perpendiculars to X-axis
Draw a perpendicular AR from the point
A on BQ.
Then OP = x1, OQ = x2
So, PQ = OQ - OP = x2 - x1
Observe the shape of APQR. It is a
rectangle.
So PQ = AR = x2 - x1.
Also QB = y2, QR = y1,
So BR = QB - QR = y2 - y1
In DARB (right triangle)
AB2
=AR2
+ RB2
(ByPythagorean theorem)
AB2
= (x2 - x1)2
+ (y2 - y1)2
i.e., AB = 2 2
2 1 2 1
( ) ( )
x x y y
- + -
Hence, ‘d’the distance between the pointsAand B is
d = 2 2
2 1 2 1
( ) ( )
x x y y
- + - .
Thisis called the distance formula.
Y
X
X¢
Y¢
R
A
(
,
)
x
y
1
1
B
(
,
)
x
y
2
2
( - )
x x
2 1
( - )
y y
2 1
O P Q
( - )
x x
2 1

Class-X Mathematics
168
Example-3. Let’s find the distance between two points A(4, 3) and B(8, 6)
Solution : Compare these points with (x1, y1), (x2, y2)
x1 = 4, x2 = 8, y1 = 3, y2 = 6
Using distanceformula
d = 2 2
2 1 2 1
( ) ( )
x x y y
- + -
distanceAB = 2 2 2 2
(8 4) (6 3) 4 3
- + - = +
16 9 25
= + = = 5 units.
DO THIS
Find thedistance betweenthe following pair of points
(i) (7, 8) and (-2, 3) (ii) (-8, 6) and (2, 0)
TRY THIS
Find the distance betweenA(1, -3) and B(-4, 4) and rounded to two decimal.
THINK - DISCUSS
Sridhar calculated the distance betweenT(5, 2) and R(-4, -1) to the nearest decimal
is 9.5 units.
Now you find the distance between P(4, 1) and Q(-5, -2). Do you get the same answer
that Sridhar got? Why?
Let us see some examples
Example-4. Showthat the pointsA(4, 2), B (7, 5) and C (9, 7) are three points lyingona same
line.
Solution : Let us find the distances AB, BC, AC by using distance formula,
2 2
2 1 2 1
( ) ( )
d x x y y
= - + -

So, AB = 2 2 2 2
(7 4) (5 2) 3 3 9 9 18
- + - = + = + =
9 2 3 2
= ´ = units.
BC = 2 2 2 2
(9 7) (7 5) 2 2 4 4 8 2 2
- + - = + = + = = units
AC = 2 2 2 2
(9 4) (7 2) 5 5 25 25 50
- + - = + = + =
25 2 5 2
= ´ = units.
Now AB + BC = 3 2 2 2 5 2
+ = =AC. Therefore, that thethree points(4, 2), (7, 5) and
(9, 7) lie ona straight line. (Points that lie onthesame line are calledcollinear points).
Example-5. Do the points (3, 2), (-2, -3) and (2, 3) forma triangle?
Solution : Let us apply the distance formula to find the lengths PQ, QR and PR, where
P(3, 2), Q(-2, -3) and R(2, 3) are the given points. We have
PQ = ( ) ( ) ( ) ( )
2 2 2 2
2 3 3 2 5 5 25 25 50
- - + - - = - + - = + = = 7.07 units (approx)
QR = ( )
( ) ( )
( ) ( ) ( )
2 2 2 2
2 2 3 3 4 6 52 7.21
- - + - - = + = = units (approx)
PR = ( ) ( ) ( )
2 2 2 2
2 3 3 2 1 1 2 1.41
- + - = - + = = units (approx)
Since the sum ofanytwo ofthese lengths is greater thanthe third length, the points P, Q
and R forma triangle andallthe sides oftriangle are unequal.
Example-6. Show that the points (1, 7), (4, 2), (-1, -1) and (-4, 4) are the vertices of a
square.
Solution : LetA(1, 7), B(4, 2), C(-1, -1)and D(-4, 4) be the given points.
One wayofshowing thatABCD is a square is to usethe propertythat allits sides should
be equaland both its diagonals should also be equal. Now, the sides are
AB = 2 2
(1 4) (7 2) 9 25 34
- + - = + = units
BC = 2 2
(4 1) (2 1) 25 9 34
+ + + = + = units
CD = 2 2
( 1 4) ( 1 4) 9 25 34
- + + - - = + = units

Class-X Mathematics
170
DA = ( ) ( )
2 2
4 1 4 7
- - + - = 25 9
+ = 34 units
and diagonalare AC = 2 2
(1 1) (7 1) 4 64 68
+ + + = + = units
BD = 2 2
(4 4) (2 4) 64 4 68
+ + - = + = units
Since AB = BC = CD = DA and AC = BD. So all the four sides of the quadrilateral
ABCD are equaland its diagonalsAC and BD are also equal. Therefore, ABCD is square.
Example-7. The figure shows the arrangement ofdesks ina class room.
Madhuri, Meena, Pallaviare seated atA(3, 1),
B(6, 4) and C(8, 6) respectively.
Do you think theyare seated in a line ?
Give reasons for your answer.
Solution : Using the distance formula, we have
A B = ( ) ( )
2 2
6 3 4 1 = 9 9 = 18 = 3 2
- + - +
units
BC= ( ) ( )
2 2
8 6 6 4 = 4 4 = 8 = 2 2
- + - +
units
AC= ( ) ( )
2 2
8 3 6 1 = 25 25 = 50 = 5 2
- + - + units
Since,AB + BC = 3 2 = 2 2 = 5 2 =AC, we can saythat the pointsA, B and C are
collinear. Therefore, they are seated ina line.
Example-8. Find the relation between xand y such that the point (x , y) is equidistant fromthe
points (7, 1) and (3, 5).
Solution : Given P(x, y) be equidistant from the pointsA(7, 1) and B(3, 5).
AP = BP. So, AP2
= BP2
i.e., (x – 7)2
+ (y – 1)2
= (x – 3)2
+ (y – 5)2
i.e., (x2
– 14x + 49) + (y2
– 2y + 1) = (x2
– 6x + 9) + (y2
– 10y + 25)
(x2
+ y2
- 14x - 2y + 50) - (x2
+ y2
- 6x - 10y + 34) = 0
1 2 3 4 5 6 7 8 9 10
10
9
8
7
6
5
4
3
2
1
A
B
C
0

-8x + 8y = -16
i.e., x – y = 2 which is the required relation.
Example-9. Find a point on the Y-axis which is equidistant from both the pointsA(6, 5) and
B(– 4, 3).
Solution : We know that a point on theY-axis is ofthe form (0, y). So, let the point
P(0, y) be equidistant fromAand B. Then
PA = ( ) ( )
2 2
6 0 5 y
- + -
PB = ( ) ( )
2 2
4 0 3 y
- - + -
PA2
= PB2
So, (6 – 0)2
+ (5 – y)2
= (– 4 – 0)2
+ (3 – y)2
i.e., 36 + 25 + y2
– 10y = 16 + 9 + y2
– 6y
i.e., 4y = 36
i.e., y = 9
So, the required point is (0, 9).
Let us checkour solution: AP = 2 2
(6 0) (5 9) 36 16 52
- + - = + =
BP = 2 2
( 4 0) (3 9) 16 36 52
- - + - = + =
So (0, 9) is equidistant from (6, 5) and (4, 3).
EXERCISE 7.1
1. Find the distance between the following pair of points
(i) (2, 3) and (4, 1) (ii) (-5, 7) and (-1, 3)
(iii) (-2, -3) and (3, 2) (iv) (a, b) and (-a, -b)
2. Find the distance between the points (0, 0) and (36, 15).
3. Verifywhether the points (1, 5), (2, 3) and (-2, -1) are collinear or not.

Class-X Mathematics
172
4. Check whether (5, -2), (6, 4) and (7, -2) are the vertices of anisosceles triangle.
5. In a class room, 4 friends are seated at the points
A, B,C and D asshownin Figure. Jarinaand Phani
walk into the class and after observing for a few
minutes Jarina asks Phani“Don’t you notice that
ABCD is a square?” Phanidisagrees.
Using distance formula, decide who is correct and
why?
6. Show that the following points forman equilateral
triangleA(a, 0), B(-a, 0), C(0, a 3 )
7. Prove that the points (-7, -3), (5, 10), (15, 8) and (3, -5) takenin order are the corners
ofa parallelogram.
8. Show that the points (-4, -7), (-1, 2), (8, 5) and (5, -4) taken in order are the vertices of
a rhombus. Find its area.
(Hint :Area ofrhombus =
1
2
´ product ofits diagonals)
9. Name the type ofquadrilateralformed, ifany, bythe following points, and give reasons for
your answer.
(i) (-1, -2), (1, 0), (-1, 2), (-3, 0) (ii) (-3, 5), (3, 1), (1, -3), (-5, 1)
(iii) (4, 5), (7, 6), (4, 3), (1, 2)
10. Find the point on the X-axis which is equidistant from(2, -5) and (-2, 9).
11. Ifthe distance between two points (x, 7) and (1, 15) is 10, find the value of x.
12. Find thevaluesofyfor whichthedistance betweenthepoints P(2, -3) and Q(10, y) is 10
units.
13. Find the radius ofthe circle whose centre is (3, 2) and passes through (-5, 6).
14. Canyou draw a triangle with vertices (1, 5), (5, 8) and (13, 14) ? Give reason.
15. Find a relation between x and y such that the point (x, y) is equidistant fromthe points
(-2, 8) and (-3, -5)
1 2 3 4 5 6 7 8 9 10
10
9
8
7
6
5
4
3
2
1
A
B
C
D

7.5 SECTION FORMULA
A andB are two towns.To reachB
fromA, wehave to travel36 kmEast and
from there15 km North of the town A
(as shown in the figure). Suppose a
telephone company wants to position a
relaytower at PbetweenAand B in such
a waythat the distance ofthe tower from
B is twice its distance fromA. IfPlies on
AB, it willdivideABinthe ratio 1 :2(See
figure). IfwetakeAasthe originO, and 1
km as one unit on both the axes, the
coordinatesofB willbe (36, 15). Inorder to know the position ofthe tower, we must know the
coordinates ofP. How do we find these coordinates?
Let the coordinates ofP be (x, y). Draw perpendiculars fromP and Bto the X-axis, meeting it in
D and E, respectively. Draw PC perpendicular to BE. Then, by the AA similarity criterion,
studied earlier, DPOD and DBPC are similar.
Therefore,
OD OP 1
PC PB 2
= = and
PD OP 1
BC PB 2
= =
So,
1
36 2
x
x
=
-
1
15 2
y
y
=
-
.
2x = (36 - x) 2y = 15 - y
3x = 36 3y = 15
x = 12 y = 5
These equations give x = 12
and y = 5.
You can check that P(12, 5)
meets the conditionthat OP: PB =
1 : 2.
Consider any two points
A(x1, y1) and B(x2, y2) and
assume that P (x, y) divides AB
internallyinthe ratio m1 : m2,
i.e., 1
2
AP
PB
m
m
= ..... (1)
(See figure).
DrawAR, PS and BT
perpendicular to the X-axis. Draw
Y
36-x
36 km
15 km
B(36, 15)
C
P
y y
x
A
36-x
D
( , )
x y
O
X
Y
R
B( )
x y
2 2
,
C
P
A
Q
( )
x y
1 1
,
O
X
S
m1
m2
( , )
x y

Class-X Mathematics
174
AQ and PC parallelto the X-axis.Then, bytheAAsimilaritycriterion,
DPAQ ~ DBPC
Therefore,
AP AQ PQ
PB PC BC
= = .....(2)
Now,AQ = RS = OS – OR = x – x1
PC = ST = OT – OS = x2 – x
PQ = PS – QS = PS – AR = y – y1
BC = BT– CT = BT – PS = y2 – y
Substituting these valuesin (1), we get
1 1 1
2 2 2
m x x y y
m x x y y
- -
= =
- -
1
2
AP
from(1)
PB
m
m
é ù
=
ê ú
ë û
Q
Taking 1 1
2 2
m x x
m x x
-
=
-
, we get 1 2 2 1
1 2
m x m x
x
m m
+
=
+
Similarly, taking 1 1
2 2
,
m y y
m y y
-
=
-
we get 1 2 2 1
1 2
m y m y
y
m m
+
=
+
So, the coordinates ofthe point P(x, y) which divides the line segment joining the points
A(x1, y1) and B(x2, y2), internally in the ratio m1 : m2 are
1 2 2 1 1 2 2 1
1 2 1 2
,
m x m x m y m y
m m m m
æ ö
+ +
ç ÷
+ +
è ø
.....(3)
This is known as the section formula.
This can also be derived by drawing perpendiculars fromA, P and B on the Y-axis and
proceeding as above.
Ifthe ratio in whichPdividesAB is k : 1, then the coordinates ofthe point Pare
2 1 2 1
, .
1 1
kx x ky y
k k
+ +
æ ö
ç ÷
+ +
è ø

Special Case : The mid-point of a line segment divides the line segment in the ratio 1 : 1.
Therefore, the coordinates ofthe mid-point Pofthe joinofthe pointsA(x1, y1) and B(x2, y2)are
1 2 1 2 1 2 1 2
1. 1. 1. 1.
, , .
1 1 1 1 2 2
x x y y x x y y
+ + + +
æ ö æ ö
=
ç ÷ ç ÷
+ +
è ø è ø
Let us solve few examples based onthe section formula.
Example-10. Find the coordinatesofthe point whichdivides the line segment joining the points
(4, -3) and (8, 5) in the ratio 3 : 1 internally.
Solution : Let P(x, y) bethe required point. Using the sectionformula
1 2 2 1
1 2 2 1
1 2 1 2
P( , ) ,
m y m y
m x m x
x y
m m m m
+
æ ö
+
= ç ÷
+ +
è ø
, we get
3(8) 1(4) 24 4 28
7,
3 1 4 4
x
+ +
= = = =
+
3(5) 1( 3) 15 3 12
3
3 1 4 4
y
+ - -
= = = =
+
P(x, y) = (7, 3) is the required point.
Example-11. Find the mid point ofthe line segment joining the points (3, 0) and (-1, 4)
Solution : The mid point M(x,y) ofthe line segment joining the points (x1, y1) and (x2, y2).
M(x, y) =
1 2 1 2
,
2 2
x x y y
+ +
æ ö
ç ÷
è ø
The mid point ofthe line segment joining the points (3, 0) and (-1, 4) is
M(x, y) =
3 ( 1) 0 4 2 4
, , (1,2)
2 2 2 2
+ - +
æ ö æ ö
= =
ç ÷ ç ÷
è ø è ø
.
DO THIS
1 Findthepoint whichdividesthelinesegment joiningthepoints(3,5)and(8,10)internally
in the ratio 2 : 3.
2. Find the midpoint ofthe line segement joining the points (2, 7) and (12, -7).

Class-X Mathematics
176
7.6 TRISECTIONAL POINTS OF A LINE
The points which divide a line segment into 3 equal parts are said to be the trisectional
points.
Example-12. Find the coordinates ofthe points oftrisection ofthe line segment joining the
pointsA(2,-2) and B(-7, 4).
Solution : Let Pand Q bethe points oftrisectionofAB i.e.,AP=PQ=QB (see figure below).
Therefore, PdividesAB internallyin the ratio 1 : 2.
Therefore, the coordinatesofP are (byapplying the sectionformula)
1 2 2 1
1 2 2 1
1 2 1 2
P( , ) ,
m y m y
m x m x
x y
m m m m
+
æ ö
+
= ç ÷
+ +
è ø
1( 7) 2(2) 1(4) 2( 2)
,
1 2 1 2
- + + -
æ ö
ç ÷
+ +
è ø
i.e., ( )
7 4 4 4 3 0
, , 1,0
3 3 3 3
- + - -
æ ö æ ö
= = -
ç ÷ ç ÷
è ø è ø
Now, Q also dividesAB internallyintheratio 2:1.
So,the coordinates of Q are
2( 7) 1(2) 2(4) 1( 2)
,
2 1 2 1
- + + -
æ ö
= ç ÷
+ +
è ø
i.e., ( )
14 2 8 2 12 6
, , 4,2
3 3 3 3
- + - -
æ ö æ ö
= = -
ç ÷ ç ÷
è ø è ø
Therefore, the coordinates ofthe points oftrisectionofthe line segment areP(-1, 0) and
Q(-4, 2)
DO THIS
1. Find the trisectionalpoints ofline joining (2, 6) and (-4, 8).
2. Find the trisectionalpointsofline joining (-3, -5) and (-6, -8).
A
(2, -2)
B
(-7, 4)
P Q

TRY THIS
LetA(4, 2), B(6, 5) and C(1, 4) be the vertices of DABC
1. AD is the median on BC. Find the coordinates ofthe point D.
2. Find the coordinates ofthe point PonAD such that AP: PD =
2 : 1.
3. Find thecoordinates ofpoints Q and R onmediansBE and CF.
4. Findthepointswhichdividethelinesegment BE intheratio 2:1andalsothat dividethe
line segment CF in the ratio 2 : 1.
5. What do you observe?
Justifythatthepointthat divideseachmedianintheratio 2:1isthecentriodofatriangle.
7.7 CENTROID OF A TRIANGLE
The centroid of a triangle is the point of
concurrencyofits medians.
LetA(x1, y1), B(x2, y2) and C(x3, y3) be the
vertices ofthe triangleABC.
Let AD be the median bisecting its base.
Then,
2 3 2 3
D ,
2 2
x x y y
+ +
æ ö
= ç ÷
è ø
Now the point G on AD which divides it internally in the ratio 2 : 1, is the centroid.
If(x, y) are the coordinates of G, then
G(x, y) =
2 3 2 3
1 1
2 1( ) 2 1( )
2 2
,
2 1 2 1
x x y y
x y
é + + ù
æ ö æ ö
+ +
ç ÷ ç ÷
ê ú
è ø è ø
ê ú
+ +
ê ú
ê ú
ë û
=
1 2 3 1 2 3
,
3 3
x x x y y y
+ + + +
é ù
ê ú
ë û
A
B C
F
D
E
P
A
B C
2
D
1
( , 1)
y
x1
( )
x y
2, 2
( )
x y
3, 3
G

Class-X Mathematics
178
Hence, the coordinates ofthe centroidare given by
1 2 3 1 2 3
,
3 3
x x x y y y
+ + + +
é ù
ê ú
ë û
.
Example-13. Find the centroid ofthe triangle whose vertices are (3, -5), (-7, 4) and (10, -2).
Solution : The coordinates ofthe centroid are
1 2 3 1 2 3
,
3 3
x x x y y y
+ + + +
æ ö
= ç ÷
è ø
Therefore, Centroid of the triangle whose vertices are (3, -5), (-7, 4) and (10, -2).
3 ( 7) 10 ( 5) 4 ( 2)
, (2, 1)
3 3
+ - + - + + -
æ ö
= -
ç ÷
è ø
The centroid is (2, -1).
DO THIS
Find thecentroidofthetrianglewhoseverticesare(-4, 6),(2, -2)and(2,5)respectively.
Example-14. In what ratio does the point (– 4, 6) divide the line segment joining the points
A(– 6, 10) and B(3, – 8)?
Solution : Let (– 4, 6) divideAB internallyinthe ratio m1 :m2. Usingthesectionformula,weget
( ) 1 2 1 2
1 2 1 2
3 6 8 10
4, 6 ,
m m m m
m m m m
æ ö
- - +
- = ç ÷
+ +
è ø
.....(1)
We know that if (x, y) = (a, b) then x = a and y = b.
So, 1 2
1 2
3 6
4
m m
m m
-
- =
+
and 1 2
1 2
8 10
6
m m
m m
- +
=
+
Now, 1 2
1 2
3 6
4
m m
m m
-
- =
+
gives us

– 4m1 – 4m2 = 3m1 – 6m2
i.e., 7m1 = 2m2
1
2
2
=
7
m
m
i.e., m1 : m2 = 2 : 7
We should verifythat the ratio satisfies the y-coordinate also.
Now,
1
1 2 2
1
1 2
2
8 10
8 10
1
m
m m m
m
m m
m
- +
- +
=
+ +
(Dividing throughout bym2)
2
8 10
7
2
1
7
- ´ +
=
+
=
16
10
7
9
7
-
+
=
16 70
9
- +
=
54
9
= 6
Therefore, the point (-4, 6) divides the line segment joining the points A(-6, 10) and
B (3, -8) in the ratio 2 : 7.
THINK - DISCUSS
The line joining pointsA(6, 9) and B(-6, -9) isgiven
(a) In whichratio does the origindivide AB ?And what is it called for AB?
(b) In which ratio does the point P(2, 3) divide AB ?
(c) Inwhichratio does the point Q(-2, -3) divide AB?
(d) In to how manyequal parts is AB divided byP and Q?
(e) What do we call P and Q for AB ?

Class-X Mathematics
180
Example-15. Find the ratio in which the y-axis divides the line segment joining the points
(5, -6) and (-1, -4).Also find the point of intersection.
Solution : Let the ratio be K : 1. Then bythe sectionformula, the coordinates ofthe point which
dividesAB in the ratio K : 1 are
K( 1) 1(5) K( 4) 1( 6)
,
K 1 K 1
- + - + -
æ ö
ç ÷
+ +
è ø
i.e.,
K 5 4K 6
,
K 1 K 1
- + - -
æ ö
ç ÷
+ +
è ø
This point lies on the Y-axis, and we know that onthe Y-axis the abscissa is 0.
Therefore,
K 5
0
K 1
- +
=
+
-K + 5 = 0 Þ K = 5.
So, the ratio is K : 1 = 5 : 1
Putting the value ofK = 5, we get the point ofintersection as
=
5 5 4(5) 6
,
5 1 5 1
- + - -
æ ö
ç ÷
+ +
è ø
=
20 6
0,
6
- -
æ ö
ç ÷
è ø
=
26
0,
6
-
æ ö
ç ÷
è ø
=
13
0,
3
-
æ ö
ç ÷
è ø
Example-16. Show that the pointsA(7, 3), B(6, 1), C(8, 2) and D(9, 4) taken inthat order are
verticesofaparallelogram.
Solution : Let the pointsA(7, 3), B(6, 1), C(8, 2) andD(9, 4) are verticesofa parallelogram.
We know that the diagonals ofa parallelogrambisect each other.
So the midpoint ofthe diagonalsAC and DB should be same.
Now, we find the mid points ofAC and DBbyusing
1 2 1 2
,
2 2
x x y y
+ +
æ ö
ç ÷
è ø
formula.
midpoint of AC =
7 8 3 2 15 5
, ,
2 2 2 2
+ +
æ ö æ ö
=
ç ÷ ç ÷
è ø è ø
midpoint of DB =
9 6 4 1 15 5
, ,
2 2 2 2
+ +
æ ö æ ö
=
ç ÷ ç ÷
è ø è ø
Hence, midpoint ofAC and midpoint of DBis same.
Therefore, the pointsA, B, C, D are vertices ofa parallelogram.
A(7,3)
C(8, 2)
B(6, 1)
D(9, 4)

Example-17. If the points A(6, 1), B(8, 2), C(9, 4) and D(p, 3) are the vertices of a
parallelogram, takeninorder, find the valueofp.
Solution :We know that diagonals ofparallelogrambisect each other.
So, the coordinates ofthe midpoint ofAC =Coordinates ofthe midpoint ofBD.
i.e.,
6 9 1 4 8 5
, ,
2 2 2 2
p
+ + +
æ ö æ ö
=
ç ÷ ç ÷
è ø è ø
15 5 8 5
, ,
2 2 2 2
p
+
æ ö æ ö
=
ç ÷ ç ÷
è ø è ø
15 8
2 2
p
+
=
15 = 8 + p Þ p = 7.
EXERCISE - 7.2
1. Find the coordinates ofthe point which divides the line segment joining the points (-1, 7)
and (4, -3) in the ratio 2 : 3.
2. Find the coordinates of the points of trisection of the line segment joining (4, -1) and
(-2, -3).
3. Findtheratio inwhichtheline segment joining thepoints(-3, 10)and (6, -8) isdivided by
(-1, 6).
4. If (1, 2), (4, y), (x, 6) and (3, 5) are the vertices of a parallelogram taken in order, find
x and y.
5. Find the coordinates of a point A, whereAB is the diameter of a circle whose centre is
(2, -3) and B is (1, 4).
6. IfAand B are (-2, -2) and (2, -4) respectively, find the coordinates of PonAB such that
AP =
3
7
AB.
7. Find the coordinatesofpoints which divide the line segment joiningA(-4, 0) and B(0, 6)
into four equalparts.

Class-X Mathematics
182
8. Find the coordinates of the points which divides the line segment joiningA(-2, 2) and
B(2, 8) into four equalparts.
9. Find the coordinates of the point which divides the line segment joining the points
(a + b, a - b) and (a - b, a + b) in the ratio 3 : 2 internally.
10. Find the coordinatesofcentroid ofthe triangles withvertices:
i. (-1, 3), (6, -3) and (-3, 6) ii. (6, 2), (0, 0) and (4, -7)
iii. (1, -1), (0, 6) and (-3, 0)
7.8 AREA OF THE TRIANGLE
Consider the points A(0, 4) and B(6, 0) which form a
triangle with originO ona plane asshownin figure.
What is the area of the DAOB?
DAOB isright angle triangle andthe base is 6units(i.e., x
coordinate) and height is 4 units (i.e., y coordinate).
Area of DAOB =
1
base height
2
´ ´
=
1
6 4 =12
2
´ ´ square units.
TRY THIS
Take a point Aon X-axis and B onY-axis and find area ofthe triangleAOB. Discuss
withyour friendshow theydo it?
THINK - DISCUSS
Let A(x1, y1), B(x2, y2), C(x3, y3) be three points.
Then find the area of the following triangles and
discuss with your friends in groups about the area of
that triangle.
X
A
Y
B
C
X'
Y'
(i)
B(6, 0)
6 units
4
units
A(0, 4)
O

Area of the triangle
Let ABC be any triangle whose
vertices are A(x1, y1), B(x2, y2) and
C(x3, y3) .
Draw AP, BQ and CR
perpendiculars from A, B and C
respectivelyto the X-axis.
ClearlyABQP,APRC and BQRC
are alltrapezia.
Now fromthe figure, it isclear that
Area ofDABC=area oftrapezium
ABQP+ area oftrapeziumAPRC - area
oftrapeziumBQRC
Q Area oftrapezium=
1
2
(sumof the parallelsides) (distance between them)
Area of DABC =
1 1 1
(BQ AP)QP (AP CR)PR (BQ CR)QR
2 2 2
+ + + - + .... (1)
Fromthe figure
BQ = y2, AP = y1, QP = OP - OQ = x1 - x2
CR = y3, PR = OR - OP = x3 - x1
QR = OR - OQ = x3 - x2
Therefore,Area of DABC [from(1)]
2 1 1 2 1 3 3 1 3 3 3 2
1 1 1
( ) ( ) ( ) ( ) ( ) ( )
2 2 2
= + - + + - - + -
y y x x y y x x y y x x
1 2 3 2 3 1 3 1 2
1
| ( ) ( ) ( ) |
2
= - + - + -
x y y x y y x y y
Y
X
B A
C
X'
Y'
(ii)
X
Y
A B
C
Y'
X'
(iii)
X
Y
B
C
A
Y'
X'
(iv)
X¢ X
Y¢
Y
C( )
x y
3 3
,
A( , )
y1
x1
B( )
x y
2 2
,
S
T
U
Q P R

Class-X Mathematics
184
Thus, the area of DABC is
1 2 3 2 3 1 3 1 2
1
( ) ( ) ( )
2
- + - + -
x y y x y y x y y
Note: As the area cannot be negative, we have takenabsolute value.
Let ustrysome examples.
Example-18. Find the area of a triangle whose vertices are (1, -1), (-4, 6) and (-3, -5).
Solution : Area ofthe triangle = D = 1 2 3 2 3 1 3 1 2
1
( ) ( ) ( )
2
- + - + -
x y y x y y x y y
The area of the triangleformed bythe verticesA(1, -1), B(-4, 6) and C(-3, -5), byusing
theformula
D
1
1(6 5) ( 4)( 5 1) ( 3)( 1 6)
2
= + + - - + + - - -
1
11 16 21 24
2
= + + =
So the areaofthe triangle is 24 square units.
Example-19. Find the area ofa triangle formed bythe pointsA(5, 2), B(4, 7) and C(7, -4).
Solution :The area ofthe triangle formed bythe vertices A(5, 2), B(4, 7) and C(7, -4) is given
by
=
1
5(7 4) 4( 4 2) 7(2 7)
2
+ + - - + -
1 4
55 24 35 2 2
2 2
-
= - - = = - =
Therefore, the areaofthe triangle = 2 square units.
DO THIS
Find the area ofthe triangle whose vertices are
1. (5, 2) (3, -5) and (-5, -1)
2. (6, -6), (3, -7) and (3, 3)

Example-20. IfA(-5, 7), B(-4,-5), C(-1, -6) and D(4,5) are the verticesofa quadrilateral,
then, find the area of the quadrilateralABCD.
Solution : Byjoining B to D, you willget two trianglesABD, and BCD.
The area of DABD
1
5( 5 5) ( 4)(5 7) 4(7 5)
2
= - - - + - - + +
106
1
50 8 48 53
2 2
= + + = = square units
Also, The area of DBCD 1
4( 6 5) 1(5 5) 4( 5 6)
2
= - - - - + + - +
1
44 10 4 19
2
= - + = Square units
So, the area ofquadrilateralABCD =Area of DABD + area of DBCD
53+19 = 72 square units.
TRY THIS
Find the area of the square formed by (0, -1), (2, 1) (0, 3) and (-2, 1) as vertices.
THINK - DISCUSS
Find the area ofthe triangle formed bythe following points
(i) (2, 0), (1, 2), (1, 6)
(ii) (3, 1), (5, 0), (1, 2)
(iii) (-1.5, 3), (6, 2), (-3, 4)
What do you observe?
Plot these points onthree different graphs. What do you observe?
Can we have a triangle with zero square units area?
What doesit mean?
A
C
B
D
C

Class-X Mathematics
186
7.8.1. COLLINEARITY
We know that the pointsthe lie on the same line are called collinear points.
Suppose the pointsA(x1, y1), B(x2, y2) and C(x3, y3) are collinear i.e. they are lying on a
line. Then, theycan not forma triangle. i.e. area of DABC is zero.
Similarly, when the area of a triangle formed bythree pointsA, B and C is zero, then three
points arecollinear points.
Example-21. The points (3, -2) (-2, 8) and (0, 4) are three pointsin a plane. Show that these
points are collinear.
Solution : Byusing area of the triangle formula
D =
1
3(8 4) ( 2)(4 ( 2)) 0(( 2) 8)
2
- + - - - + - -
1
12 12 0
2
= - =
The areaof the triangle is 0. Hence the three points are collinear i.e., theylie onthe same
line.
DO THIS
Verifywhether thefollowing pointsare collinear
(i) (1, -1), (4, 1), (-2, -3)
(ii) (1, -1), (2, 3), (2, 0)
(iii) (1, -6), (3, -4), (4, -3)
7.8.2. AREA OF A TRIANGLE- ‘HERON’S FORMULA’
We know theformula for area ofthetriangle is
1
base height
2
´ ´ .
Anygiventriangle maybe aright angle triangle,
equilateral triangle or an isosceles triangle. How do
we calculate its area?
If we know the base and height directly, we
applythe above formula to find the area ofa triangle.
However,iftheheight (h) isnot known,how do
we find its area?

For this Heron, anAncient Greek mathematician, derived a formula for a triangle whose
lengths ofsides a, b and c are known. The formula is:
A ( )( )( )
s s a s b s c
= - - - , where s
2
a b c
+ +
=
For example, wefind the area ofthe triangle whose lengthsofsides are 12m, 9m, 15m by
using Heron’sformula we get
A ( )( )( )
s s a s b s c
= - - - , where
2
a b c
s
+ +
=
s
12 9 15 36
18
2 2
m
+ +
= = =
Then s - a = 18 - 12 = 6m
s - b = 18 - 9 = 9m
s - c = 18 - 15 = 3m
A 18(6)(9)(3) 2916 54
= = = square meters.
DO THIS
(i) Find the area ofthe triangle the lengths ofwhose sides are 7m, 24m, 25m(use
Heron’s Formula).
(ii) Find the area of the triangle formed by the points (0, 0), (4, 0), (4, 3) by using
Heron’sformula.
Example-22. Find the value of ‘b’for which the pointsA(1, 2), B(-1, b) and C(-3, -4) are
collinear.
Solution : Let the given pointsA(1, 2), B(-1, b) and C(-3, -4)
Then x1 = 1, y1 = 2; x2 = -1, y2 = b; x3 = -3, y3 = -4
We know, area of D = 1 2 3 2 3 1 3 1 2
1
( ) ( ) ( )
2
- + - + -
x y y x y y x y y
areaofDABC=
1
1( 4) ( 1)( 4 2) ( 3)(2 ) 0
2
b b
+ + - - - + - - = (Q The givenpoints are collinear)
|b + 4 + 6 - 6 + 3b| = 0
|4b + 4| = 0
4b + 4 = 0
b = -1

Class-X Mathematics
188
EXERCISE - 7.3
1. Find the area ofthe triangle whose vertices are
(i) (2, 3) (-1, 0), (2, -4) (ii) (-5, -1), (3, -5), (5, 2)
(iii) (0, 0), (3, 0) and (0, 2)
2. Find the valueof‘K’for which thepoints are collinear.
(i) (7, -2) (5, 1) (3, K) (ii) (8, 1), (K, -4), (2, -5)
(iii) (K, K) (2, 3) and (4, -1).
3. Find the area of the triangle formed byjoining the mid-points ofthe sides ofthe triangle
whose vertices are (0, -1), (2, 1) and (0, 3). Find the ratio ofthis area to the area ofthe
giventriangle.
4. Find the area ofthe quadrilateral whose vertices, takenin order, are (-4, -2), (-3, -5),
(3, -2) and (2, 3).
5. Find the areaofthetriangleformedbythepoints (2, 3), (6, 3) and (2, 6) byusing Heron’s
formula.
7.9 STRAIGHT LINES
Bharadwajand Meenaare discussing to find solutions for a linear equationin two variable.
Bharadwaj : Can you find solutions for 2x + 3y = 12
Meena :Yes, I have found some of them. x 0 3 6 -3
y 4 2 0 6
Ingeneral, 2x+3y = 12
3y = 12 - 2x
y =
12 2
3
x
-
Meena : Can you write these solutions
in ordered pairs?
Bharadwaj : Yes, (0, 4), (3, 2), (6, 0),
(-3, 6)
Meena, canyou plot these points
on the coordinate plane?
Meena : I have done like this.
Bharadwaj : What do you observe?
-9
Y
O
X
1
Y
1
X
-8 -7 -6 -5 -4 -3 -2 -1 1 2 3 4 5 6 7 8 9
-1
-2
-3
-4
-5
-6
-7
-8
-9
1
2
3
4
5
6
7
8
9
A(0,4)
B(3,2)
C(6,0)
D(-3,6)
Scale

What does this figure represent?
Meena : It is a straight line.
Bharadwaj : Canyou identifysome more pointson thisline?
Can you help Meena to find some more points onthis line?
..................., ..................., ..................., ...................
And in this line, what is AB called ?
AB is a line segment.
DO THIS
Plot these points onthe coordinate planeand join them:
1. A(1, 2), B(-3, 4), C(7, -1)
2. P(3, -5) Q(5, -1), R(2, 1), S(1, 2)
Which one is a straight line? Whichis not? Why?
THINK - DISCUSS
Does y = x + 7 represent a straight line? Draw the line on the coordinate plane.
At whichpoint doesthis lineintersectY-axis?
How muchangle doesit make withX-axis? Discuss withyour friends
7.9.1 SLOPE OF THE STRAIGHT LINE
You might have seen a slide in a park. Two slides have been given here. On which slide
you canslide faster?
Obviously your answer will be second. Why?
Observe these lines.

Class-X Mathematics
190
l
X
O O X
m
Which line makes more angle withOX ?
Since the line “m” makes a greater angle withOX thanline ‘l’,
line ‘m’has a greater “slope”than line ‘l’. We mayalso termthe “Steepness” ofa line as
its slope.
How canwe measure the slope ofa line?
ACTIVITY
Consider the linegivenin the figureindentifythe pointsonthe line and fillthe table below.
x coordinate 1 - - 4 -
y coordinate 2 3 4 - 6
We canobserve that ycoordinates
change when x coordinates change.
When ycoordinateincreasesfrom
y1 = 2 to y2 = 3,
So the change in y is = ........................
Thencorrespondingchange in‘x’is= ...

change in
change in
y
x
= ..................
When y coordinate increases from
y1 = 2 to y3 = 4,
the change in y is = ..................
the corresponding change inxis............
So,
change in
change in
y
x
= ...............
-9
Y
O
X
1
Y
1
X
-8 -7 -6 -5 -4 -3 -2 -1 1 2 3 4 5 6 7 8 9
-1
-2
-3
-4
-5
-6
-7
-8
-9
1
2
3
4
5
6
7
8
9
Scale

Then, can you tryother points onthe line? Choose anytwo points and fillin the table.
y value Change in y x Change in x
change in
change in
y
x
2 4 - 1 2 1 -
- - - - - - -
- - - - - - -
What do you conclude fromabove activity?
Therefore, there is a relation between the ratio ofchange in y to change in x on aline has
relationwithangle made byit with X-axis.
You willlearntheconcept oftanqfromtrigonometry
i.e.,
Oppositesideof angle
tan
adjacentsideof angle
q
q =
q
Change in
Change in
=
y
x
7.9.2 SLOPE OF A LINE JOINING TWO POINTS
Let A(x1, y1) and B(x2, y2) be two points on a line ‘l’not parallel to Y-axis as shown in
figure.
The slope of a line =
change in
change in
y
x
Slope of AB = m =
2 1
2 1
-
-
y y
x x
Slope will be denoted by‘m’ and the line
‘l’ makes the angle qwith X-axis.
SoAB line segment makes the same angle q
withAC also.

Oppositesideof angle
tan
adjacentsideof angle
q
q =
q
= 2 1
2 1
BC
AC
-
=
-
y y
x x
2 1
2 1
tan
y y
m
x x
-
q = =
-
Hence m = 2 1
2 1
tan
-
q =
-
y y
x x
It is theformula to find slope ofline segment AB whichishaving end pointsare (x1,y1),
(x2, y2).
If q is angle made bythe line with X-axis, then m =tan q.
Y
X
X¢
Y¢
C
A
(
,
)
y
1
x
1
B
(
)
x
y
2
2
,
x x
2 1
-
y y
2 1
-
q

Class-X Mathematics
192
Example-22. Theendpointsofalinesegment are(2,3),(4,5). Findtheslopeofthelinesegment.
Solution : The end points ofthe line segment are (2, 3), (4, 5), the slope of the line segment is
2 1
2 1
5 3 2
1
4 2 2
- -
= = = =
- -
y y
m
x x
Slope ofthe givenline segment is 1.
DO THIS
Find the slope of AB with the given end points.
1. A(4, -6), B(7, 2) 2. A(8, -4), B(-4, 8)
3. A(-2, -5), B(1, -7)
TRY THIS
Find the slope of AB
suu
r
, where
1. A(2, 1), B(2, 6) 2. A(-4, 2), B(-4, -2)
3. A(-2, 8), B(-2, -2)
4. Justify that the line AB
suu
r
line segment formed bypoints given in the above three
examples isparallelto Y-axis. What can you say about their slope?Why?
THINK - DISCUSS
Find the slope of AB
suu
r
passing throughA(3, 2) and B(-8, 2)
Is the line AB
suu
r
parallelto X-axis? Why?
Think and discuss with your friends ingroups.
Example-23. Determine x so that 2 is the slope of the line passing throughP(2, 5) and Q(x,3).
Solution : Slope ofthe line passing through P(2, 5) and Q(x, 3) is 2.
Here, x1 = 2, y1 = 5, x2 = x, y2 = 3
Slope of 2 1
2 1
3 5 2 2
PQ 2
2 2 2
- - - -
= = = Þ =
- - - -
y y
x x x x x
Þ -2 = 2x - 4 Þ 2x = 2 Þ x = 1

EXERCISE - 7.4
1. Find the slope ofthe linepassing throughthe giventwo points
(i) (4, -8) and (5, -2)
(ii) (0, 0) and ( 3,3)
(iii) (2a, 3b) and (a, -b)
(iv) (a, 0) and (0, b)
(v) A(-1.4, -3.7), B(-2.4, 1.3)
(vi) A(3, -2), B(-6, -2)
(vii)
1 1
A 3 , 3 , B 7, 2
2 2
æ ö æ ö
- -
ç ÷ ç ÷
è ø è ø
(viii) A(0, 4), B(4, 0)
OPTIONAL EXERCISE
1. Centre of a circle Q is on the Y-axis. The circle passes through the points (0, 7) and
(0, -1). Ifit intersects the positive X-axis at (P, 0), what is the value of ‘P’?
2. Atriangle DABC is formed bythe pointsA(2, 3), B(-2, -3),C(4, -3). What isthe point
ofintersectionofthe side BC and the bisector ofangleA?
3. The side BC ofan equilateraltriangle DABC is parallelto X-axis. Find the slopesofthe
lines along sides BC, CAandAB.
4. Find the centroid of the triangle formed by the line 2x + 3y - 6 = 0, withthe coordinate
axes.
Suggested Projects
l Byusing graph paper find the coordinates of apoint which divides a givenline segment
internally.

Class-X Mathematics
194
1. The distance between two points P(x1, y1) and Q(x2, y2) is ( ) ( )
2 2
2 1 2 1
x x y y
- + - .
2. The distance ofa point P(x, y) fromthe origin is 2 2
x y
+ .
3. The distance between two points (x1, y1)and(x1, y2) ona lineparalleltoY-axis is |y2 - y1|.
4. The distance between two point (x1, y1) and (x2, y1) on a line parallelto X-axis is |x2 - x1|.
5. The coordinates of the point P(x, y) which divides the line segment joining the points
A (x1, y1) and B(x2, y2) internally in the ratio m1 : m2 are
1 2 2 1 1 2 2 1
1 2 1 2
,
m x m x m y m y
m m m m
é ù
+ +
ê ú
+ +
ë û
.
6. The mid-point of the line segment joining the points (x1, y1) and (x2, y2) is
1 2 1 2
,
2 2
x x y y
+ +
æ ö
ç ÷
è ø
.
7. The centroid ofa triangle isthe point of intersectionofits medians. Hencethe coordinates
of the centroid are
1 2 3 1 2 3
,
3 3
x x x y y y
+ + + +
æ ö
ç ÷
è ø
, where (x1, y1) (x2, y2) and (x3, y3) are
the verticesofthe triangle.
8. The point that divides each median ofa trianglein the ratio 2 : 1 is the centroid.
9. The area ofthe triangle formed bythe points (x1, y1), (x2, y2) and (x3, y3) isthe numerical
value of the expression
1
2
|x1 (y2 - y3) + x2 (y3 - y1) + x3 (y1 - y2)|
10. Area of a triangle is given by‘Heron’s Formula’as
A ( )( )( )
s s a s b s c
= - - - , where
2
a b c
s
+ +
=
(a, b, c are three sides of DABC)
11. Slope of the line containing the points (x1, y1) and (x2, y2) is m =
2 1
2 1
y y
x x
-
- ( x1 ¹ x2)

8.1 INTRODUCTION
There is a tall tree in the
backyard of Snigdha’s house.
She wants to find out the height
of that tree but she is not sure
about howtofindit.Meanwhile,
her uncle arrives at home.
Snigdha requests her uncle to
help her with the height. He
thinks for a while and then asks her to bring a mirror. He places it on the ground at a certain
distance fromthe base of thetree. He thenasked Snigdha to stand on the otherside ofthe mirror
at such a position fromwhere she is able to see the top ofthe tree inthat mirror.
When we draw the figure from(AB) girlto the mirror (C) and mirror to the tree (DE) as
above, weobserve trianglesABC and DEC. Now, what can you sayabout these two triangles?
Are theycongruent? No, because although theyhave the same shape their sizesare different.
Do you know what we call the geometrical figures which have the same shape, but are not
necessarilyofthe same size? Theyare called similarfigures.
Canyou guess how the heights oftrees, mountains or distances offar-away, objects such
as the Sun have beenfound out? Do you think these can bemeasured directlywiththehelp ofa
measuringtape?Thefactisthatalltheseheights anddistanceshavebeenfoundout usingtheidea
of indirect measurements whichis based on the principle ofsimilarityof figures.
8.2 SIMILAR FIGURES
Observe the object (car) in the figure (i).
If its breadth is kept the same and the lengthis doubled, it appears as infig.(ii).
A
B C
D
E
(i) (ii) (iii)
Similar Triangles
8

Class-X Mathematics
196
Ifthe lengthin fig.(i) is kept the same and its breadthis doubled, it appearsas in fig.(iii).
Now, what canyou sayabout fig.(ii) and (iii)? Do theyresemble fig.(i)? We find that the
figure isdistorted. Canyou saythat they aresimilar? No, theyhave same shape, yet they are not
similar.
Think what a photographer does when she prints photographs ofdifferent sizes fromthe
same film(negative) ? You might have heard about stamp size, passport size andpost card size
photographs. She generally takes a photograph on a small size film, say 35 mm., and then
enlargesit into a bigger size, say45 mm (or55 mm). We observe that everyline segment ofthe
smaller photographisenlarged in theratio of35 : 45 (or 35 : 55). Further, inthetwo photographs
ofdifferent sizes, we can see that the corresponding angles are equal. So, the photographs are
similar.
Similarly in geometry, two polygons of the same number of sides are similar if their
corresponding angles areequaland their corresponding sides are in the sameratio or proportion.
A polygon in which all sides and angles are equal is called a regular polygon.
Theratio ofthecorrespondingsidesisreferredto asscalefactor(orrepresentativefactor).
In reallife, blue prints for theconstruction ofa building are prepared using a suitable scale factor.
THINK AND DISCUSS
Can you give some more examples from your dailylife where scale factor is used.
All regular polygonshaving the same number of sides are always similar. For example, all
squares are similar, allequilateral triangles are similar and so on.
Circles with same radius are
congruent andthosewithdifferent radiiare
not congruent. But, asallcircleshavesame
shape, theyare allsimilar.
Wecansaythat allcongruent figures
are similar but allsimilar figures need not
be congruent.
Similar
Squares
Similar equilateral
triangles
Similar
Circles
(i) (ii) (iii)

Similar Triangles 197
To understand thesimilarityoffigures moreclearly, let us performthe following activity.
ACTIVITY
Suspendatransparentplasticsheet horizontallyfromtheceilingof
the roof. Fixa lighted bulb at the point ofsuspension. Then, a shadow of
quadrilateralABCDiscast onthetable. Marktheoutlineoftheshadow
as quadrilateral A¢ B¢ C¢ D¢.
Nowthisquadrilateral A¢ B¢ C¢ D¢ isenlargementormagnification
of quadrilateralABCD. Further, A¢ lies on ray OAwhere ‘O’is the
bulb, B¢on OB
uuu
r
, C¢on OC
uuu
r
and D¢on OD
uuu
r
. QuadrilateralsABCD and A¢ B¢ C¢ D¢are ofthe
same shape but of different sizes.
A¢ corresponds to vertexAand we denote it symbolicallyas A¢ « A. Similarly B¢ «B,
C¢ « C and D¢ «D.
Byactuallymeasuring angles and sides, you can verify
(i) A = A¢
Ð Ð , B = B¢
Ð Ð , C = C¢
Ð Ð , D = D¢
Ð Ð and
(ii) AB BC CD DA
A B B C C D D A
= = =
¢ ¢ ¢ ¢ ¢ ¢ ¢ ¢
.
Thisemphasises that two polygons withthesame number ofsidesare similar if
(i) Allthe corresponding anglesareequaland
(ii) Allthe lengths ofthe correspondingsides are in the same ratio (or inproportion)
Is a square similar to a rectangle? In both the figures, corresponding angles are equalbut
their corresponding sidesare not in the same ratio. Hence, theyare not similar. For similarityof
polygons onlyone ofthe above two conditions is not sufficient, both have to be satisfied.
THINK AND DISSUSS
Can you say that a square and a rhombus are similar? Discuss with your friends.
Write whytheconditionsare not sufficient.
A B
C
D
¢
A

Class-X Mathematics
198
DO THIS
1. Fillinthe blanks withsimilar/ not similar.
(i) Allsquaresare ........................
(ii) Allequilateraltriangles are ........................
(iii) Allisosceles trianglesare ........................
(iv) Two polygonswithsamenumberofsidesare ........................ iftheir corresponding
angles areequaland corresponding sides are equal.
(v) Reduced and Enlarged photographs of anobject are ........................
(vi) Rhombus and squares are ........................ to eachother.
2. WriteTrue/ Falsefor thefollowing statements.
(i) Anytwo similarfiguresarecongruent.
(ii) Anytwo congruent figures are similar.
(iii) Two polygons aresimilar iftheir correspondingangles are equal.
3. Give two different examples of pair of
(i) Similarfgures (ii)Nonsimilar figures
8.3 SIMILARITY OF TRIANGLES
In the example of finding a tree’s height bySnigdha, we had drawn two triangles which
showed the propertyof similarity. We knowthat, two trianglesare similar iftheir
(i) Corresponding angles are equaland
(ii) Lengthsofthe corresponding sides are in the same ratio (inproportion)
In DABCand DDEC in the introduction,
A
Ð = D
Ð , B
Ð = E
Ð , ACB = DCE
Ð Ð
Also
DE EC DC
= = = K
AB BC AC
(scale factor)
thus DABC is similar to DDEC.
Symbolically we write DABC ~ DDEC
(Symbol‘~’is read as “Issimilar to”)
As we have stated that K is a scale factor, So
if K > 1, we get enlarged figures,
K= 1, we get congruent figures and
K < 1, we get reduced (or diminished) figures
A
B C
D
E

Further, intrianglesABC and DEC, corresponding angles are
equal. So theyare called equiangulartriangles. The ratio ofanytwo
corresponding sides intwo equiangular triangles is always the same.
For proving this, Basic Proportionalitytheoremis used. Thisis also
knownasThales Theorem.
To understandBasicproportionalitytheoremorThalestheorem,
let usdo the followingactivity.
ACTIVITY
Take anyruled paper and draw a triangle on it with base on
one of the lines. Several lines will cut the triangleABC. Select
anyone line among themand name the points whereit meets the
sides AB and AC as P and Q.
Find the ratio of
AP
PB
and
AQ
QC
. What do you observe?
The ratios will be equal. Why ? Is it always true? Try for different lines intersecting the
triangle. We know that allthe lines on a ruled paper are paralleland we observe that every
time the ratios are equal.
So in DABC, if PQ || BC then
AP
PB
=
AQ
QC
.
This is knownas the result ofbasic proportionalitytheorem.
8.3.1 BASIC PROPORTIONALITY THEOREM (THALES THEOREM)
Theorem-8.1 : If a line is drawn parallelto one side ofa triangle to intersect the other two sides
in distinct points, then the other two sides are divided inthe same ratio.
Given : In DABC, DE || BC, and DE intersects sidesAB andAC at D and E respectively.
RTP:
AD AE
=
DB EC
Construction : Join B, E and C, D and then draw
DM ^ AC and EN ^ AB.
Proof : Area of DADE =
1
AD EN
2
´ ´
Area of DBDE =
1
BD EN
2
´ ´
A
P Q
B C
Basic proportionality
theorem?
A
N M
D E
B C

Class-X Mathematics
200
So
1
AD EN
ar( ADE) AD
2
1
ar( BDE) BD
BD EN
2
´ ´
D
= =
D ´ ´
...(1)
AgainArea ofDADE =
1
AE DM
2
´ ´
Area of DCDE =
1
EC DM
2
´ ´
1
AE DM
ar( ADE) AE
2
1
ar( CDE) EC
EC DM
2
´ ´
D
= =
D ´ ´
...(2)
Observe that DBDE and DCDE areon the same base DE and between sameparallels BC
and DE.
So ar(DBDE) = ar(DCDE) ...(3)
From (1) (2) and (3), we have
AD AE
=
DB EC
Hence proved.
Is the converse of the above theorem also true? To examine this, let us perform the
followingactivity.
ACTIVITY
Draw an angle XAYinyour note book and on rayAX, mark points B1, B2, B3, B4 and
B whichare equidistant respectively.
AB1 = B1B2 = B2B3 = B3B4 = B4B = 1cm(say)
Similarlyon rayAY, mark points C1, C2, C3, C4 and C such that
AC1 = C1C2 = C2C3 = C3C4 = C4C = 2 cm (say)
Join B1, C1 and B, C.
Observe that
1 1
1 1
AB AC 1
=
B B C C 4
=

Similarly, joining B2C2, B3C3 and B4C4, you see that
2 2
2 2
AB AC 2
=
B B C C 3
= and
3 3
3 3
AB AC 3
=
B B C C 2
= and
4 4
4 4
AB AC 4
=
B B C C 1
= and
check whether C1B1 || C2B2 || C3B3 || C4 B4 || CB?
Fromthiswe obtain the following theorem called converse ofthe Thales theorem
Theorem-8.2 : If a line divides two sides ofa triangle in the sameratio, then the lineis parallelto
the third side.
Given : In DABC, a line DE is drawnsuchthat
AD AE
DB EC
=
RTP: DE || BC
Proof : Assume that DE is not parallel to BC then draw the line
DE1
parallelto BC
So
AD AE
DB E C
¢
=
¢
(why ?)
AE AE
EC E C
¢
=
¢
(why ?)
Adding 1 to both sides ofthe above, you can see that E and E¢ must coincide (why?)
TRY THIS
1. In triangle DPQR, E and F are pointson the sides PQ and PR respectively. For each of
the following,state whether EF||QR or not?
(i) PE = 3.9 cm EQ = 3 cm PF = 3.6 cm and FR = 2.4 cm
(ii) PE = 4 cm, QE = 4.5 cm, PF = 8 cm and RF = 9 cm.
(iii) PQ = 1.28 cm PR = 2.56 cm PE = 1.8 cm and PF = 3.6 cm
Y
C1
A B1
C2
C3
C
C4
B2 B3 B4 B
X
A
D E
E
1
B C

Class-X Mathematics
202
2. In the followingfigures DE || BC.
(i) Find EC (ii) FindAD
Construction : Division of a line segment (using Thales theorem)
Madhuridrew aline segment. She wantsto divide it
in the ratio of3 : 2. She measured it byusing a scale and
divided it inthe required ratio. Meanwhileher elder sister
came. She saw this and suggested Madhuri to divide the
linesegment inthegivenratio without measuringit Madhuri
waspuzzled and asked her sister forhelp to do it. Thenher
sisterexplained. Youmayalsodo it bythefollowingactivity.
ACTIVITY
Take a sheet of paper from a lined note book.
Number the lines by 1, 2, 3, ... starting with the bottom
linenumbered‘0’.
Take a thick cardboard paper (or file card or chart
strip) and placeit against the givenline segmentAB
andtransferitslengthtothecard. LetA1
andB1
denote
the points onthe file card corresponding toAand B.
Now place A1
on the zeroeth line of the lined
paper and rotate the card about A1
unitl point B1
falls on the 5th
line (3 + 2).
Mark the point where the third line touches the file card, byP1
.
Again place this card along the given line segment andtransfer this point P1
and denote
it with‘P’.
So ‘P’is the required point which divides the givenline segment inthe ratio 3:2.
Nowlet uslearnhow this constructioncanbe done.
Given a line segmentAB. We want to divide it in
the ratio m: n where mand n arebothpositive integers.
Let us take m = 3 and n = 2.
Steps :
1. Draw a rayAX throughAmaking an acute angle
withAB.
0
9
8
7
6
5
4
3
2
1
A1
P1
B1
A
1
B1
A B
A
B C
D E
7
.
2
.
c
m
5
.
4
.
c
m
1
.
8
.
c
m
1
.
5
.
c
m
A
B C
D
E
3
.
c
m
1
.
c
m
A B
X

2. With ‘A’as centre and with anylength draw an arc
on rayAX and label the point A1.
3. Usingthe
same compass
setting and with
A1 as centre draw another arc and locateA2.
4. Like this locate 5 points (=m + n) A1, A2, A3, A4, A5
such that AA1 = A1A2 = A2A3 = A3A4 =A4A5
5. JoinA5B. Now through point A3(m = 3) draw a line parallel to
A5B (bymaking anangle equalto A
Ð A5B) intersectingAB at C
and observe that AC : CB = 3 : 2.
Now let us solve someexamplesusing Thales theoremand its converse.
Example-1. In DABC, DE || BC and
AD 3
.
DB 5
=
AC = 5.6. Find AE.
Solution : In DABC, DE || BC
AD AE
DB EC
Þ = (byB.P.T)
but
AD 3
DB 5
= So
AE 3
EC 5
=
GivenAC = 5.6 andAE : EC = 3 : 5.
AE 3
AC AE 5
=
-
AE 3
5.6 AE 5
-
(crossmultiplication)
5AE = (3 ´5.6) - 3AE
8AE = 16.8
AE =
16.8
2.1cm.
8
=
A
D E
B C
A B
X
A1
A B
X
2
C
3
A1
A2
A3
A4
A5
A B
X
A1
A2
A3
A4
A5

Class-X Mathematics
204
Example-2. In the given figure LM ||AB
AL = x - 3, AC = 2x, BM = x - 2
and BC = 2x + 3 find the value of x
Solution : In DABC, LM ||AB
AL BM
LC MC
Þ = (byB.P.T)
3 2
2 ( 3) (2 3) ( 2)
x x
x x x x
- -
=
- - + - -
3 2
3 5
x x
x x
- -
+ +
(crossmultiplication)
(x - 3) (x + 5) = (x - 2) (x + 3)
x2
+ 2x - 15 = x2
+ x - 6
2x - x = - 6 + 15
x = 9
DO THIS
1. What value(s) of x willmake DE ||AB, in the given figure ?
AD = 8x + 9, CD = x + 3
BE = 3x + 4, CE = x.
2. In DABC, DE || BC. AD = x, DB = x - 2,
AE = x + 2 and EC = x - 1.
Find thevalue ofx.
Example-3. The diagonals ofa quadrilateralABCD intersect each other at point ‘O’such that
AO
BO
=
CO
DO
. Prove thatABCD isa trapezium.
Solution : Given : In quadrilateralABCD ,
AO CO
=
BO DO
.
RTP:ABCDisa trapezium.
Construction : Through‘O’draw a line paralleltoAB which meetsDAat X.
Proof : In DDAB, XO || AB (byconstruction)
Þ
DX
XA
=
DO
OB
(bybasic proportionalitytheorem)
A
M
L
B
C
A D
E
B
C
E
C
D
B
A

AX
XD
=
BO
OD
..... (1)
again
AO
BO
=
CO
DO
(given)
AO
CO
=
BO
OD
..... (2)
From (1) and (2)
AX
XD
=
AO
CO
In DADC, XO is a line such that
AX
XD
=
AO
OC
Þ XO || DC (by converse of the basic the proportionality theorem)
Þ AB || DC
In quadrilateralABCD,AB || DC
Þ ABCD isa trapezium (bydefinition)
Hence proved.
Example-4. IntrapeziumABCD,AB || DC. E and F are points on non-parallelsidesAD and
BC respectivelysuch that EF ||AB. Show that
AE BF
=
ED FC
.
Solution : Let us joinA, C to intersect EF at G.
AB || DC and EF ||AB (given)
Þ EF || DC(Lines parallelto the same line are parallelto each other)
In DADC, EG || DC
So
AE AG
=
ED GC
(byBPT) ...(1)
Similarly, In DCAB, GF ||AB
CG CF
=
GA FB
(by BPT) i.e.,
AG BF
=
GC FC
...(2)
From (1) & (2) AE BF
=
ED FC
.
A
D
G
E
B
F
C
D
A
O
X
C
B

Class-X Mathematics
206
EXERCISE - 8.1
1. In DPQR, ST is a line such that
PS PT
=
SQ TR
and
also PST PRQ
Ð = Ð .
Prove that DPQR is an isosceles triangle.
2. In the given figure, LM || CB and LN || CD
Prove that
AM AN
=
AB AD
3. In the given figure, DE ||AC and DF||AE
Prove that
BF BE
=
FE EC
.
4. Prove that a line drawn throughthe mid-point ofone side ofa triangle parallelto another
side bisectsthe third side (Using basic proportionalitytheorem).
5. Prove that aline joining the midpointsofanytwo sides ofa triangle is parallelto the third
side. (Using converse ofbasic proportionalitytheorem)
6. In thegiven figure, DE ||OQ and DF || OR. Show
that EF || QR.
7. In the adjacent figure,A, B and C are points on OP, OQ
and OR respectivelysuch thatAB || PQ andAC||PR.
Show that BC || QR.
P
S T
Q R
P
D
O
E
Q R
F
P
A
B
Q R
C
O
A
D
E
C
B
F
A L
C
N
D
M
B

8. ABCD isa trapeziuminwhichAB||DC and its diagonalsintersect eachother at point ‘O’.
Show that AO CO
=
BO DO
.
9. Draw alinesegment oflength7.2cmanddivideit intheratio5 :3. Measurethetwo parts.
THINK ABD DISCUSS
Discuss with your friends that in what waysimilarityof triangles is different from
similarityofother polygons?
8.4 CRITERIA FOR SIMILARITY OF TRIANGLES
Weknowthat two trianglesaresimilarifcorresponding anglesareequalandcorresponding
sides are proportional. For checking the similarity of two triangles, we should check for the
equalityofcorresponding angles andequalityofratios oftheir corresponding sides. Let usmake
anattempt toarrive at certaincriteriafor similarityoftwo triangles. Let usperformthe following
activity.
ACTIVITY
Use a protractor and ruler to draw two non congruent triangles so that each triangle
should have 40°and 60°angle. Check the figures madebyyou bymeasuringthe third angles
oftwo triangles.
It should be each 80° (why?)
Measure the lengths ofthe sides ofthe triangles and computethe ratios ofthelengths ofthe
corresponding sides.
Are thetrianglessimilar?
Thisactivityleadsusto thefollowing criterionfor similarityoftwo triangles.
8.4.1 AAA CRITERION FOR SIMILARITY OF TRIANGLES
Theorem-8.3 : In two triangles, if corresponding angles are equal, then their corresponding
sides are inproportion and hence the trianglesare similar.
Given : IntrianglesABC and DEF,
A
Ð = D
Ð , B
Ð = E
Ð and C
Ð = F
Ð
A
B C
60
°
40
°
P
Q R
60°
40°

Class-X Mathematics
208
RTP :
AB BC AC
DE EF DF
= =
Construction : If AB<DE andAC<DF, locate points P and Q on DE and DF
respectively, such that AB = DP andAC = DQ. Join PQ.
Proof : ABC DPQ
D @ D (why?)
This gives B
Ð = P
Ð = E
Ð and PQ || EF (How ?)
DP DQ
PE QF
= (why?)
i.e.,
AB AC
DE DF
= (why ?)
Similarly
AB BC
DE EF
= and So
AB BC AC
.
DE EF DF
= =
Hence proved.
In the above construction, ifAB=DE orAB>DE, what will you do? Just think!
Note : If two angles ofa triangle are respectivelyequalto the two anglesofanother triangle, then
bythe angle sumpropertyofa triangle, third angleswillalso be equal.
SoAAsimilaritycriterionisstatedasiftwo anglesofonetrianglearerespectivelyequalto
the two angles ofanther triangle, then the two triangles are similar.
What about the converse of the above statement?
If the sides ofa triangle are respectivelyproportional to the sides of another triangle, is it
true that theircorresponding angles are equal?Let us exercise it throughan activity.
ACTIVITY
Draw two triangles ABC and DEF such that AB = 3 cm, BC = 6 cm, CA = 8 cm,
DE = 4.5 cm, EF = 9 cm and FD = 12 cm.
So you have
AB BC CA 2
.
DE EF FD 3
= = =
Now measure the angles ofboththe triangles. What do you observe?What can you say
about the corresponding angles? Theyare equal, so the triangles are similar. You can verify
it fordifferent triangles.
Fromtheaboveactivity,wecanwritethefollowingcriterionforsimilarityoftwo triangles.
D
E F
A
B C
A
B C
D
E F
P Q

8.4.2. SSS Criterion for Similarity of Triangles
Theorem-8.4: Intwotriangles, ifcorrespondingsidesareproportional, thentheir corresponding
angles are equaland hence the triangles are similar.
Given : DABC and DDEF are such that
AB BC CA
( 1)
DE EF FD
= = <
RTP : A
Ð = D
Ð , B
Ð = E
Ð , C
Ð = F
Ð
Construction : Let DE >AB Locate points Pand Q on DE and DF respectively
such that AB = DP and AC = DQ. Join PQ.
Proof :
DP DQ
PE QF
= and PQ || EF (why ?)
So P
Ð = E
Ð and Q
Ð = F
Ð (why ?)
DP DQ PQ
DE DF EF
= =
So
DP DQ BC
DE DF EF
= = (why ?)
So BC = PQ (Why ?)
ABC DPQ
D @ D (why ?)
So A
Ð = D
Ð , B
Ð = E
Ð and C
Ð = F
Ð (How ?)
We studied that for similarityoftwo polygons anyonecondition is not sufficient. But for
thesimilarityoftriangles, thereisno needforfulfillmentofboththeconditionsasoneautomatically
implies the other. Now let us look for SAS similarity criterion. For this, let us perform the
followingactivity.
A
B C
D
E F
P Q

Class-X Mathematics
210
ACTIVITY
Draw two trianglesABC and DEF such that AB = 2 cm, A
Ð =500
AC = 4cm,
DE = 3 cm, D
Ð = 500
and DF = 6cm.
Observe that
AB AC 2
DE DF 3
= = and A
Ð = D
Ð = 500
.
Now measure B
Ð , C
Ð , E
Ð , F
Ð also measure BC and EF.
.
Observe that B
Ð = E
Ð and C
Ð = F
Ð also
BC 2
.
EF 3
=
So,thetwotrianglesaresimilar. Repeat thesamefortriangleswithdifferent measurements,
whichgivesthe following criterionforsimilarityoftriangles.
8.4.3 SAS CRITERION FOR SIMILARITY OF TRIANGLES
Theorem-8.5 : If one angle of a triangle is equal to one angle of the other triangle and the
including sides ofthese angles areproportional, then the two triangles are similar.
Given : In DABC and DDEF
AB AC
( 1)
DE DF
= < and
A
Ð = D
Ð
RTP : DABC ~ DDEF
Construction : Locate points P and Q on DE and DF respectively such that AB = DP and
AC = DQ. Join PQ.
Proof : PQ || EF and ABC DPQ
D @ D (How ?)
So A
Ð = D
Ð , B
Ð = P
Ð , C
Ð = Q
Ð
DABC ~ DDEF (why ?)
A
B C
D
E F
P Q
A
B C
D
E F

TRY THIS
1. Are trianglesformed ineachfiguresimilar? Ifso,name the criterionofsimilarity. Write
the similarityrelationinsymbolicform.
(i) (ii)
(iii) (iv)
(v) (vi)
(vii) (viii)
2. If pairs ofthe triangles are similar and then find the value of x.
(i) (ii)
(iii) (iv)
(v) (vi)
K
G
H
I
F P
Q R
L
M N
A
B C
X Y
3
2 2
3
A
B C
P J
5
3 2
31
3
Q
O
P
B
A
A
B C
40°
60° 80°
P
Q R
60°
40° 80°
3 cm
.
2 cm.
2
cm.
6
cm
.
5
cm.
4 cm.
70°
5 cm.
2
.5
c
m
.
6
c
m
.
10 cm.
70°
3
5
x
4.5
110°
70°
3 3
5
x
x
22
24 14
2
x
9
6
x
5
4
15 18
7.5
x
12
A
B C
P
Q
R
A
B C
P
Q R
S
L
T
P
Q
R
A
B
C
E
D
A
B C
P
Q
Z
X
Y
P
Q R
A
B
C
M N
A B
S T
R
x

Class-X Mathematics
212
(vii) (viii)
Construction : To construct a triangle similar to a given triangle as per given scale factor.
a) Constructatrianglesimilarto agiventriangleABCwithitssidesequalto
3
4
ofcorresponding
sides of DABC (scale factor
3
4
)
Steps : 1. Draw a rayBX, making an acute angle with BC on the side
opposite to vertexA.
2. Locate 4 points B1, B2, B3 and B4 on BX so that BB1 = B1B2 =
B2B3 = B3B4.
3. Join B4C and drawa line through B3
parallel to B4C intersectingBC at C¢.
4. Draw a line through C¢ parallelto CAto intersectAB at A¢.
So A BC
D ¢ ¢ isthe required triangle.
Let ustake some examples to illustrate the use ofthese criteria.
Example-5. Aperson 1.65mtall casts 1.8mshadow. At the same instance, a lamp post casts
a shadow of5.4 m. Find the height ofthe lamppost.
Solution: In DABC and DPQR
B
Ð = Q
Ð = 900
.
C
Ð = R
Ð (AC || PR, all sun’s rays are parallel at any instance)
DABC ~ DPQR ( byAAsimilarity)
AB BC
PQ QR
= (cpst,corresponding parts of Similar triangles)
A
B C
C
1
B1
B2
B3
B4
X
A
1
1.8 m.
B
A
1.65 m.
h m.
Q 5.4 m. R
P
C
A
C
D
B 4cm E x
3cm
5cm
A
B
C
E
D
1.5 cm 15cm
x
1.6
cm

1.65 1.8
PQ 5.4
=
1.65 5.4
PQ 4.95m
1.8
´
= =
The height ofthe lamp post is 4.95m.
Example-6. Amanseesthe top ofa tower in a mirrorwhich is at a distance of87.6m fromthe
tower. The mirror is on the ground facing upwards. The man is 0.4m away from the
mirror andhis height is 1.5m. How tallisthe tower?
Solution : In DABC & DEDC
ABC EDC 90
Ð = Ð = °
BCA DCE
Ð = Ð (Complements of the
angle of incidence and angle of reflection
arecongruent)
DABC ~ DEDC (byAAsimilarity)
AB BC 1.5 0.4
ED CD h 87.6
= Þ =
1.5 87.6
h 328.5m
0.4
´
= =
Hence, the height of the towers is 328.5m.
Example7. Gopalisworryingthat hisneighbour canpeepinto his livingroomfromthetopfloor
of his house. He has decided raise the height of the fencethat is highenoughto block the
view fromhisneighbour’stop floor window. What should bethe height ofthefence? The
measurements are givenin the figure.
Solution : In DABD & DACE
B
Ð = C
Ð = 90°
A
Ð = A
Ð (common angle)
DABD ~ DACE (byAAsimilarity)
AB BD 2 BD
AC CE 8 1.2
= Þ =
2 1.2 2.4
BD 0.3m
8 8
´
= = =
Totalheight ofthe fence required is 1.5 m. +0.3 m. = 1.8mto block the neighbour’sview.
E
0.4 m.C
h
B
1.5 m.
87.6 m Tower
D
Mirror
A
1.5 m.
R Q
C
B A
2m.
8 m.
P
1.5 m. 1.5 m.
D
E
1.2 m.

Class-X Mathematics
214
EXERCISE - 8.2
1. In thegivenfigure, ADE B
Ð = Ð
(i) Show that DABC ~ DADE
(ii) IfAD = 3.8 cm, AE = 3.6cm,
BE = 2.1 cm and BC = 4.2 cm,
findDE.
2. The perimeters oftwo similar triangles are 30 cmand 20 cmrespectively. If one side of
the first triangle is 12 cm, determine the corresponding side ofthe second triangle.
3. In the givenfigure, AB || CD || EF.
given that AB=7.5 cm, DC= y cm
EF = 4.5 cm and BC = x cm, find
the values of x and y.
4. A girl of height 90 cm is walking away from the base of a lamp post at a speed of
1.2 m/sec. Ifthe lamp post is 3.6mabove the ground, find the lengthofher shadow after
4 seconds.
5. Given that DABC ~ DPQR, CM and
RN are respectively the medians of
DABC and DPQR. Prove that
(i) DAMC ~ DPNR
(ii)
CM AB
RN PQ
=
(iii) DCMB ~ DRNQ
6. DiagonalsAC and BD of a trapeziumABCD withAB || DC intersect each other at the
point ‘O’. Using the criterionofsimilarityfor two triangles,show that
OA OB
OC OD
= .
7. AB, CD, PQ are perpendicular to BD.
If AB = x, CD = y and PQ = z
prove that
1 1 1
+ =
x y z
.
B
Q
z
P
A
D
C
y
A
D
F
B
C
E
7
.
5
c
m
x cm
3 cm
4
.
5
c
m
y cm
M
B
A
N
Q
R
P
A
E
B C
D

8. Aflagpole 4mtallcastsa 6 mshadow. At the sametime, a nearbybuildingcastsa shadow
of24m. How tallis the building ?
9. CD and GH are respectively the bisectors of ACB
Ð and EGF
Ð such that D and H lie
on sidesAB andFE of DABC andDFEG respectively. IfDABC ~ DFEG, thenshow that
(i)
CD AC
GH FG
= (ii) DDCB ~ DHGE (iii) DDCA~ DHGF
10. AX and DYare altitudes oftwo similar triangles DABC andDDEF. Provethat AX :DY=
AB : DE.
11. Construct atrianglesimilartothegiven DABC,withitssidesequalto
5
3
ofthecorresponding
sides ofthe triangleABC.
12. Construct a triangle ofsides 4cm, 5 cmand 6 cm. Then, construct a triangle similar to it,
whose sides are
2
3
ofthe corresponding sides of the first triangle.
13. Construct anisoscelestrianglewhosebaseis8cmandaltitudeis4cm. Then,drawanother
triangle whose sides are 1
1
2
times the corresponding sidesoftheisosceles triangle.
8.5 AREAS OF SIMILAR TRIANGLES
For two similar triangles, ratios of
their corresponding sides is the same. Do
you think thereisanyrelationship between
the ratios of their areas and the ratios of
their corresponding sides ? Let us do the
following activityto understand this.
ACTIVITY
Make a list of pairs of similar
polygonsinthisfigure.
Find
(i) the ratio ofsimilarity (scale factor)
and
(ii) the ratioofareas.
You willobserve that ratio ofareas is the square of the ratio oftheir corresponding sides.
Let us prove it like a theorem.

Class-X Mathematics
216
Theorem-8.6 : The ratio ofthe areas
oftwo similar triangles is equalto the
ratio of the squares of their
Given : DABC ~ DPQR
RTP :
2 2 2
ar( ABC) AB BC CA
ar( PQR) PQ QR RP
D æ ö æ ö æ ö
= = =ç ÷
ç ÷ ç ÷
D è ø
è ø è ø
.
Construction : DrawAM ^ BC and PN ^ QR.
Proof :
1
BC AM
ar( ABC) BC AM
2
1
ar( PQR) QR PN
QR PN
2
´ ´
D ´
= =
D ´
´ ´
...(1)
In DABM & DPQN
B
Ð = Q
Ð (Q DABC ~ DPQR)
0
M N 90
Ð = Ð =
DABM ~ DPQN (byAAsimilarity)
AM AB
PN PQ
= ...(2)
Also DABC ~ DPQR (given)
AB BC AC
PQ QR PR
= = ...(3)
ar( ABC) AB AB
ar( PQR) PQ PQ
D
= ´
D
from(1), (2) and (3)
2
AB
PQ
æ ö
= ç ÷
è ø
.
Now byusing (3), we get
2 2 2
ar( ABC) AB BC AC
ar( PQR) PQ QR PR
D æ ö æ ö æ ö
= = =ç ÷
ç ÷ ç ÷
D è ø
è ø è ø
Hence proved.
Now let us see some examples.
M
B
A
N
Q R
P

Example-8. Prove that ifthe areas oftwo similar triangles are equal, thentheyare congruent.
Solution : DABC ~ DPQR
So
2 2 2
ar( ABC) AB BC AC
ar( PQR) PQ QR PR
D æ ö æ ö æ ö
= = = ç ÷
ç ÷ ç ÷
D è ø
è ø è ø
But
ar( ABC)
1
ar( PQR)
D
=
D
(Q areas are equal)
2 2 2
AB BC AC
1
PQ QR PR
æ ö æ ö æ ö
= = =
ç ÷
ç ÷ ç ÷
è ø
è ø è ø
So AB2
= PQ2
BC2
= QR2
AC2
= PR2
From which we getAB = PQ
BC = QR
AC = PR
DABC @ DPQR (bySSS congruency)
Example-9. DABC ~ DDEF and their areas are respectively 64cm2
and 121 cm2
.
IfEF = 15.4 cm., then find BC.
Solution :
2
ar( ABC) BC
ar( DEF) EF
D æ ö
= ç ÷
D è ø
2
64 BC
121 15.4
æ ö
= ç ÷
è ø
8 BC 8 15.4
BC 11.2cm
11 15.4 11
´
= Þ = = .
Example-10. Diagonals ofa trapeziumABCD withAB ||DC, intersect each other at the point
‘O’. IfAB = 2CD, find the ratio ofareas oftrianglesAOB and COD.
Solution : In trapeziumABCD, AB || DC also AB = 2CD.
In DAOB and DCOD
AOB
Ð = COD
Ð (verticallyopposite angles)
OAB OCD
Ð = Ð (alternateinterior angles) A B
D C
O

Class-X Mathematics
218
DAOB ~ DCOD (byAAsimilarity)
2
2
ar( AOB) AB
ar( COD) DC
D
=
D
2
2
(2DC) 4
1
(DC)
= =
ar(DAOB) : ar(DCOD) = 4 : 1.
EXERCISE - 8.3
1. D, E, Fare mid points ofsides BC, CA,AB of DABC. Find the ratio ofareas of DDEF
and DABC.
2. In DABC, XY || AC and XYdivides the triangle into two parts of equal area. Find the
ratio of
AX
XB
.
3. Prove that the ratio ofareas oftwo similar triangles is equalto the square of the ratio of
their corresponding medians.
4. DABC ~ DDEF. BC = 3cm, EF= 4cmand area of DABC = 54 cm2
. Determine the area
ofDDEF.
5. ABC is atriangle and PQ isa straight line meetingAB inPandACin Q. If AP= 1 cm,
BP = 3cm, AQ = 1.5 cm and CQ = 4.5 cm, prove that area of DAPQ =
1
16
(area of
DABC).
6. The areas oftwo similar triangles are 81cm2
and49cm2
respectively. Ifthe altitude ofthe
bigger triangle is4.5 cm. Find thecorresponding altitude of the smaller triangle.
8.6 PYTHAGORAS THEOREM
You are familarwiththePythagorastheorem. Youhad verifiedthistheoremthroughsome
activities. Nowwe shallprove this theoremusing the concept ofsimilarityoftriangles. For this,
we make use ofthe following result.
Theorem-8.7 : Ifaperpendicular is drawn fromthe vertex of theright angle ofaright triangle to
the hypotenuse, then the triangles onboth sides of the perpendicular are similar to the whole
triangle and to each other.

C
A
B
C
A
B
Proof:ABCisaright triangle, rightangledat B. LetBD betheperpendicularto hypotenuseAC.
In DADB and DABC
A
Ð = A
Ð
and ADB ABC
Ð = Ð (why?)
So DADB ~ DABC (how?) ...(1)
Similarly, DBDC ~ DABC (how?) ...(2)
So from (1) and (2), triangles on both sides of the perpendicular BD are similar to the
triangleABC.
Also since DADB ~ DABC
DBDC ~ DABC
So DADB ~ DBDC (Transitive Property)
Thisleadsto the following theorem.
THINK AND DISCUSS
For a right angled trianglewith integer sides atleast one of its measurements must be
an evennumber. Why? Discuss thiswith your friendsand teachers.
8.6.1 PYTHAGORAS THEOREM (BAUDHAYAN THEOREM)
Theorem-8.8 : Ina right triangle, the square of length of the hypotenuse is equalto the sum of
the squaresoflengths ofthe other two sides.
Given: DABC is a right triangle right angled at B.
RTP : AC2
= AB2
+ BC2
Construction : Draw BD^ AC.
Proof : DADB ~ DABC
AD AB
AB AC
Þ = (sides areproportional)
AD . AC = AB2
...(1)
Also, DBDC ~ DABC
CD BC
BC AC
Þ =
CD . AC = BC2
...(2)

Class-X Mathematics
220
On adding (1) & (2)
AD . AC + CD . AC = AB2
+ BC2
AC (AD + CD) = AB2
+ BC2
AC . AC = AB2
+ BC2
2 2 2
AC AB BC
= +
The above theorem was earlier given by an ancient Indian mathematician Baudhayan
(about 800 BC)inthe following form.
“The diagonalofa rectangle producesbyitselfthe same area as produced byits both sides
(i.e. length and breadth).” So sometimes, this theorem is also referred to as the Baudhayan
theorem.
What about the converse of the above theorem?
We prove it like a theorem, as done earlier also.
Theorem-8.9 : Ina triangle ifsquare ofthe length ofone side is equalto the sumofsquares of
the lengthsofthe other two sides, then the angle opposite to the first side is a right angle and the
triangle isaright angled triangle.
Given : In DABC,
AC2
= AB2
+ BC2
RTP : B
Ð = 900
.
Construction:Constructaright
angled triangle DPQR right
angled at Q such that PQ =AB
and QR = BC.
Proof : In DPQR, PR2
= PQ2
+ QR2
(Pythagorean theorem as Q
Ð = 900
)
PR2
=AB2
+ BC2
(byconstruction) ...(1)
butAC2
=AB2
+ BC2
(given) ...(2)
AC = PR from (1) & (2)
Now In DABC and DPQR
AB = PQ (byconstruction)
BC = QE (byconstruction)
AC = PR (proved)
ABC PQR
D @ D (bySSS congruency)
B
Ð = Q
Ð (by cpct)
Q
P
R
B
C
A

but Q
Ð = 90° (by construction)
B
Ð = 90°.
Hence proved.
Now let us take some examples.
Example-11. Aladder 25mlongreachesawindowofbuilding20mabovetheground. Determine
the distance fromthe foot ofthe ladder to the building.
Solution : In DABC, C
Ð = 90°
2 2 2
AB AC BC
Þ = + (byPythagoreantheorem)
252
= 202
+ BC2
BC2
= 625 - 400 = 225
BC 225 15m
= =
Hence, the foot ofthe ladder is ata distance of15mfromthe building.
Example-12. BLand CM are medians ofa triangleABC right angled atA.
Prove that 4(BL2
+ CM2
) = 5BC2
.
Solution : BL and CM are medians of DABC in which A
Ð = 90°.
In DABC
BC2
=AB2
+AC2
(Pythagorean theorem ...(1)
In DABL, BL2
= AL2
+AB2
So
2
2 2
AC
BL AB
2
æ ö
= +
ç ÷
è ø
(Q Listhe midpoint ofAC)
2
2 2
AC
BL AB
4
= +
4BL2
= AC2
+ 4AB2
...(2)
In DCMA, CM2
= AC2
+ AM2
CM2
=
2
2 AB
AC
2
æ ö
+ ç ÷
è ø
(QM is the mid point ofAB)
A
C
B
20 m.
25 m.
A
C
B
M
L

Class-X Mathematics
222
CM2
=
2
2 AB
AC
4
+
4CM2
= 4AC2
+ AB2
...(3)
On adding (2) and (3), we get
4(BL2
+ CM2
) = 5(AC2
+ AB2
)
4(BL2
+ CM2
) = 5BC2
from (1).
Example-13. ‘O’is anypoint inside a rectangleABCD.
Prove that OB2
+ OD2
= OA2
+ OC2
Solution : Through ‘O’draw PQ || BC so that P lies onAB and Q lies on DC.
Now PQ || BC
PQ ^ AB & PQ ^ DC (Q B
Ð = C
Ð = 90°)
So, BPQ 90
Ð = ° & CQP 90
Ð = °
BPQC andAPQD are both rectangles.
Now from DOPB, OB2
= BP2
+ OP2
...(1)
Similarly from DOQD, we have OD2
= OQ2
+ DQ2
...(2)
From DOQC, we have OC2
= OQ2
+ CQ2
...(3)
and from DOAP, OA2
= AP2
+ OP2
Adding (1) & (2)
OB2
+ OD2
= BP2
+ OP2
+ OQ2
+ DQ2
= CQ2
+ OP2
+ OQ2
+ AP2
(QBP = CQ and DQ = AP)
= CQ2
+ OQ2
+ OP2
+ AP2
= OC2
+ OA2
(from (3) & (4))
DO THIS
1. In DACB, C
Ð = 900
and CD ^ AB
AB
Prove that
2
2
BC BD
.
AD
AC
=
2. A ladder 15mlong reaches a window which is 9 m above the ground on one side ofa
street. Keeping its foot at thesame point, the ladder isturned to other side ofthe street
to reacha window 12mhigh. Find the widthofthe street.
B
A
C
A D
Q
C
B
O
P

3. Inthe given fig. ifAD ^ BC
Prove that AB2
+ CD2
= BD2
+AC2
.
Example-14. The hypotenuse ofa right triangle is 6mmore thantwice ofthe shortest side. If
the third side is 2 m., less than the hypotenuse, find the sides ofthe triangle.
Solution : Let the shortest side be x m.
Then hypotenuse = (2x + 6)m and third side= (2x + 4)m.
ByPythagores theorem, we have
(2x + 6)2
= x2
+ (2x + 4)2
4x2
+ 24x + 36 = x2
+ 4x2
+ 16x + 16
x2
- 8x - 20 = 0
(x - 10) (x + 2) = 0
x = 10 or x = -2
But x can’t be negative as it is a side ofa triangle.
x = 10
Hence, the sidesofthe triangle are 10m, 26mand 24m.
Example-15. ABC is a right triangle right angled at C. Let BC = a, CA= b,AB = c and let p
be the length ofperpendicular fromC onAB. Prove that (i) pc = ab (ii) 2 2 2
1 1 1
p a b
= + .
Solution :
(i) CD ^ AB and CD = p.
Area of DABC =
1
AB CD
2
´ ´
=
1
2
cp .
also area of DABC =
1
BC AC
2
´ ´
=
1
2
ab
1 1
2 2
cp ab
=
cp = ab ...(1)
a
D
p
b
A
C
B
c
D
A
C
B

Class-X Mathematics
224
(ii) Since DABC isa right triangle right angled at C.
AB2
= BC2
+ AC2
c2
= a2
+ b2
2
2 2
æ ö
= +
ç ÷
è ø
ab
a b
p
2 2
2 2 2 2 2
1 1 1
.
+
= = +
a b
p a b a b
EXERCISE - 8.4
1. Provethatthesumofthesquaresofthesidesofarhombus
is equalto the sumofthe squares ofits diagonals.
2. ABC is a right triangle right angledat B. Let D and E be
anypoints onAB and BC respectively.
Prove that AE2
+ CD2
=AC2
+ DE2
.
3. Prove that three times the square of any side of an
equilateral triangleis equalto four times the square ofthe altitude.
4. PQR is a triangle right angled at P and M is a point on QR such that PM ^ QR .
Show that PM2
= QM × MR.
5. ABD is a triangle right angled atAandAC ^ BD
Show that (i) AB2
= BC × BD.
(ii) AC2
= BC × DC
(iii)AD2
= BD × CD.
6. ABC is anisosceles triangle right angled at C. Prove that AB2
= 2AC2
.
7. ‘O’is anypoint inthe interior ofa triangleABC.
If OD ^ BC, OE ^ AC and OF ^ AB, show that
(i) OA2
+OB2
+OC2
-OD2
-OE2
-OF2
= AF2
+BD2
+CE2
(ii) AF2
+ BD2
+ CE2
= AE2
+ CD2
+ BF2
.
8. Awireattachedto averticalpoleofheight 18mis24mlongandhasastakeattachedto the
other end. How far fromthe base of the pole should the stake be driven so that the wire
willbe taut?
9. Two poles ofheights 6m and 11mstand on a plane ground. Ifthe distance between the
feet ofthe poles is 12mfind the distance between their tops.
C
E
B
D
A
B
C
A
B D
C
F
A
E
O

10. InanequilateraltriangleABC, Dis apoint onside BC such that BD =
1
3
BC. Prove that
9AD2
= 7AB2
.
11. Inthegivenfigure,ABCisatriangleright angledat
B. D and E are ponts on BC trisect it.
Prove that 8AE2
= 3AC2
+ 5AD2
.
12. ABCisanisoscelestrianglerightangled
at B. Similar trianglesACD andABE
are constructed on sidesAC andAB.
Find the ratio between the areas of
DABE and DACD.
13. Equilateraltrianglesare drawn onthe three sidesofaright angled triangle. Show that the
area ofthetriangle on the hypotenuse is equalto the sumofthe areas oftriangles on the
other two sides.
14. Prove that the area of the equilateraltriangle described on the side of a squareis halfthe
areaoftheequilateraltriangles described on its diagonal.
8.7 DIFFERENT FORMS OF THEORITICAL STATEMENTS
1. Negation of a statement :
We have a statement and ifwe add “Not” after thestatement, we willget a new statement;
which is callednegation of the statement.
For example take a statement “DABC is a equilateral”. If we denote it by “p”, we can
write like this.
p :TriangleABC is equilateralandits negationwillbe“TriangleABCisnot equilateral”.
Negation of statement p is denoted by ~p; and read as negation of p. the statement ~p
negates the assertion that the statement p makes.
When we write the negation ofthe statements we would be careful that there should no
confusion;inunderstanding thestatement.
Observe thisexamplecarefully
P:Allirrationalnumbers arerealnumbers.We canwritenegationof p like this.
C
E
B D
A
D
C
A
B
E

Class-X Mathematics
226
i) ~p :Allirrationalnumbers are not realnumbers.
How do wedecide this negation is trueor false?We use the following criterion“Let p be
a statement and ~p itsnegation. Then~pisfalsewhenever p istrue and ~p istruewhenever
p is false.
For example s : 2 + 2 = 4 is True
~ s : 2 + 2 ¹ 4 is False
2. Converse of a statement :
A sentence whichis either true or false is called a simple statement. Ifwe combine two
simplestatementsthenwewillgetacompoundstatement.Connectingtwosimplestatements
with the use of the words “Ifand then” willgive a compound statement which is called
implication(or)conditional.
Combining two simple statements p&q usingifandthen, weget pimplies q whichcanbe
denoted by p Þ q. In this p Þ q, suppose we interchange p and q we get q Þ p. This
is called its converse.
Example : p Þ q : In DABC, ifAB =AC then C B
Ð = Ð
Converse q Þ p : In DABC, if C B
Ð = Ð thenAB =AC
3. Proof by contradiction :
In this proofbycontradiction, we assume the negation ofthe statement astrue;which we
have to prove. In the process of proving we get contradiction somewhere. Then, we
realizethat thiscontradictionoccursbecause ofour wrong assumptionwhichisnegationis
true. Therefore weconclude that the originalstatement is true.
OPTIONAL EXERCISE
1. Inthegivenfigure,
QT QR
PR QS
= and 1 2
Ð = Ð
prove that DPQS ~ DTQR. 2
1
Q
R
P
T

2. Raviis1.82mtall. Hewantsto find the height
ofa tree in his backyard. Fromthe tree’s base
he walked 12.20 m. along the tree’s shadow
to a position where the end of his shadow
exactly overlapsthe end ofthetree’s shadow.
He is now6.10mfromthe endofthe shadow.
How tallisthe tree ?
3. The diagonalAC of a parallelogramABCD intersects DP at the point Q, where ‘P’isany
point on sideAB. Prove that CQ´PQ = QA´QD.
4. DABC and DAMP are two right triangles right
angled at B and M respectively.
Prove that (i) DABC ~ DAMP and
(ii)
CA BC
PA MP
= .
5. An aeroplane leaves an airport and flies due north
at a speed of1000 kmph.At the same time, another aeroplane leaves the same airport and
flies due west at a speed of 1200 kmph. How far apart will the two planes be after
1
1
2
hour?
6. Inaright triangleABCrightangledat C,PandQarepointsonsidesACandCBrespectively
which divide these sides in the ratio of2 : 1.
Prove that (i) 9AQ2
= 9AC2
+ 4BC2
(ii) 9BP2
= 9BC2
+ 4AC2
(iii) 9(AQ2
+ BP2
) = 13AB2
1. Two figures havingthe same shape are called similar figures.
2. All thecongruent figures are similar but the converse is not true.
C
M
A
B
P
B
A
E
D
C
12.20 m.
6.10 m.
1.82 m.
Suggested Projects
l Find theheight ofa tree/tower/ temple etc. using the properties ofsimilar triangles, use
the procedure discussed in ‘Introduction ofSimilar Triangles’chapter.

Class-X Mathematics
228
3. Two polygons ofthe same number ofsides are similar,
if (i) their corresponding angles are equaland
(ii) their corresponding sides are inthe same ratio (i.e. proportion)
4. Ifa line is drawn parallelto one side ofa triangle to intersect the other two sidesat distinct
points thenthe other two sides are divided in the same ratio.
5. Ifa line divides anytwo sides ofa triangle in the same ratio, then the line isparallelto the
third side.
6. Intwo triangles,ifcorrespondinganglesareequal,thentheircorrespondingsidesare inthe
same ratio and hence the two triangles are similar (AAA similarity)
7. Iftwo anglesofa triangle are equalto the two angles of another triangle, then third angles
of bothtriangles are equalbyangle sumpropertyof triangle.
8. In two triangles, if corresponding sides are in the same ratio, then their corresponding
anglesare equaland hence the triangles aresimilar. (SSS similar)
9. If one angleofa triangle isequalto one angle ofanother triangle and theincluding sides of
these angles arein the same ratio, then the triangles aresimilar. (SAS similarity)
10. The ratio of areas of two similar triangles is equal to the square of the ratio of their
11. If a perpendicular is drawn fromthe vertex ofthe right angle to the hypotenuse in a right
angle triangle,thenthetrianglesformedonbothsidesoftheperpendicular aresimilarto the
whole triangle and also to each other.
12. In a right triangle, the square ofthe hypotenuse is equalto the sum of the squares of the
other two sides (PythagoreanTheorem).
13. Inatriangle, ifsquare ofone side is equalto the sumofthe squares oftheother two sides,
thenthe angle opposite to the first side isa right angle.
Puzzle
Draw a triangle. Jointhe mid-point of the sides of the triangle. You get 4 triangles again
join the mid-points of these triangles. Repeat this process. All the triangles drawn are
similar triangles. Why?Think anddiscusswith your friends.

9.1 INTRODUCTION
We have seen that two lines in a plane may intersect at a point or may not intersect. In
some situationstheymaycoincide witheach other.
Similarly, what are the possiblerelative positionsof a curve and a line given ina plane?A
curve maybe a parabola as you have seen in polynomialsor a simple closed curve like a “circle”
which is a collection of allthose points on a plane that are at a constant distance froma fixed
point.
You might have seen circular objects rolling on a plane creating a path. For example;
bicycle wheelon a sandy field, wheels of train on the track etc., where a circle as wellas a line
are involved. Doesthere exist a relation between them?
Let us observe the relative positionsof a circle and a line are givenin a plane.
9.1.1 A LINE AND A CIRCLE
You are asked to draw acircle and a line on a paper. Abhiramargues that there can only
be 3 possible ways of presenting themon a paper.
Consider a circle with centre ‘O’ and a line PQ. The three possibilities are given in the
followingfigures.
O
P
Q
(i)
P
O
Q
A
B
(ii)
O
P
Q
(iii)
A
Tangents and Secants
to a Circle
9
Y
X
O

Class-X Mathematics
230
In Fig.(i), the line PQ and the circle have no common point. In this case PQ is a non-
intersecting line withrespect to the circle.
In Fig.(ii), the line PQ intersects the circle at two pointsAand B. It forms a chord AB
with its end pointsAand B on the circle. In this case the line PQ is a secant of the circle.
InFig.(iii), there is onlyone pointA, commonto the line PQ and the circle. This line is
called a tangent to the circle.
You can see that there cannot be any other position ofthe line with respect to the circle.
Wewillstudytheexistenceoftangentsto acircleandalso studytheirpropertiesand constructions.
Do you know?
The word ‘tangent’comes fromthe latinword ‘tangere’, whichmeans to touch and was
introduced byDanish mathematicianThomasFinekein 1583.
DO THIS
i. Draw acircle withanyradius. Draw four tangentsat different
points. How manymore tangentscanyou draw to this circle?
ii. How many tangents can you draw to a circle from a point
awayfromit?
iii. In the adjacent figure, which lines are tangentsto the circle?
9.2 TANGENTS OF A CIRCLE
We can see that tangent to a circle can be drawn at any point on the circle. How many
tangents can be drawn at anypoint on the circle?
To understand this let us consider the following activity.
ACTIVITY
Take a circular wireand attacha straight wireAB at a point P of
the circular wire, so that the systemcan rotate about the point P in a
plane.Thecircularwirerepresentsacircle
and thestraight wireABrepresentsa line
intersects the circle at point P.
Place the systemon a table and gentlyrotate thewireAB about
thepoint Ptogetdifferent positionsofthestraight wire as shown
in the figure. The wire intersects thecircular wire at P and at one
of the points Q1, Q2 or Q3 etc. So while it generally intersects
circular wire at two points one of which is P in one particular
position, it intersects the circle onlyat the point P (See position
O
A
P
B
l
p
q
m
O
B
A
P
Q
1
Q
2
Q
3
A
11
A
1
B
11
B
1
O

Tangents and Secants to a Circle 231
A B
¢ ¢ ofAB). Thisisthepositionofatangentat thepoint Pto thecircle.Youcancheckthat
inallotherpositionsofABit willintersectthecircleat Pand atanother point.Thus A B
¢ ¢isa
tangent to the circle at P.
We see that there is onlyone tangent to the circleat point P.
MovingwireABineither directionfromthispositionmakesit cut thecircular wireintwo
points.Allthese are therefore secants. Tangent is a specialcase of a secant where the two
points of intersectionof a line witha circle coincide.
DO THIS
DrawacircleandasecantPQto thecircleonapaperasshownin
thefigure. Drawvariouslinesparallelto thesecant onbothsidesofit.
What happens to the length ofchord coming closer and closer
to the centreofthe circle?
What is the longest chord?
How manytangents canyou drawto a circle, whichare parallel
to each other ?
The common point ofa tangent and the circle is called the point ofcontact and the tangent
is said to touch the circle at the commonpoint.
Observe the tangentsto the circle inthefigures givenbelow:
How manytangents can you draw to a circle at apoint on it? How manytangents can
you draw to the circle inall?See the pointsofcontact. Draw radiifromthe pointsofcontact. Do
you see anything special about the angle between the tangents and the radii at the points of
contact.Allappear to be perpendicularto the corresponding tangents.We canalso prove it. Let
us see how.
Theorem-9.1 : The tangent at any point of a circle is
perpendicular to the radius through the point ofcontact.
Given : A circle with centre ‘O’ and a tangent XY to the
circle at a point P and OP radius.
To prove : OPis perpendicular to XY.
O
Y
A
P
A
P
Q
O

Class-X Mathematics
232
Proof : Take a point Q on XY
suur
other than Pand joinO and
Q. The point Q must lie outside the circle (why?) (Note that
if Q lies inside the circle, XY becomes a secant and not a
tangent tothecircle)
Therefore, OQislongerthantheradiusOPofthecircle
[Why?]
i.e., OQ > OP.
This must happenfor allpointsonthe line XY.It is therefore truethat OPis the shortest of
allthe distancesofthe point O to the XY.
As aperpendicular is the shortest inlength among allline segments drawnfroma point to
the line (Activity5.3 of 7thclass). Therefore OP is perpendicular to XY.
i.e., OP ^ XY
Hence proved.
Note : The linecontaining theradius throughthe point ofcontact is also called the ‘normalto
the circleat the point’.
TRY THIS
How can you prove the converse ofthe above theorem.
“Ifalineintheplaneofacircleisperpendicularto theradiusat itsendpointonthecircle,
then theline is tangent to the circle”.
We can find some more results using the above theorem
(i) Since there canbe onlyone perpendicular OP at the point P, it follows that one and only
one tangent canbe drawn to a circle at a givenpoint on the circle.
(ii) Since there can be only one perpendicular to XY at the point P, it follows that the
perpendicular to a tangent to a circle at its point ofcontact passes throughthe centre.
Think about these. Discuss these among your friends and withyour teachers.
9.2.1 CONSTRUCTION OF TANGENT TO A CIRCLE
Howcanweconstruct alinethat wouldbetangentto acircleat agivenpoint onit?Weuse
what we just found i.e. the tangent has to be perpendicular to the radius at the point ofcontact.
To draw a tangent throughthe point ofcontact we need to draw a line prependicular to the radius
at that point.To draw this radiuswe need to knowthe center ofthe circle. Let us see the steps for
thisconstruction.
O
Y
A
Q
P

Construction : Construct atangent to a circleat a given point onit, whenthe centreofthe circle
isknown.
We have a circle with centre ‘O’and a point Panywhere onits circumference. Then we have to
construct atangent throughP.
Let us observe steps ofconstructionto draw a tangent.
Steps of Construction :
1. Drawacirclewithcentre‘O’andmark
a point ‘P’anywhere on it. JoinO and P.
2. Draw a perpendicular line throughthe
point P and name it as XY, as shown in
thefigure.
3. XYis the required tangent to the givencirclepassing throughP.
Canyou drawone more tangent through P?give reason.
TRY THIS
How canyoudraw the tangent to a circleat a givenpoint whenthe
centre ofthe circle is not known?
Hint : Draw equal angles ÐQPX and PRQ
Ð . Explain the
construction.
9.2.2 FINDING LENGTH OF THE TANGENT
Canwe find the lengthofthe tangent to a circle froma givenpoint?
Example : Find the length ofthe tangent to a circle with centre ‘O’ and radius = 6 cm. from a
point P such that OP = 10 cm.
Solution :Tangent is perpendicular to the radius at the point ofcontact (Theorem9.1)
Here PAistangent segment and OAisradius ofcircle
OA PA OAP 90
^ Þ Ð = °
Now in DOAP, OP2
= OA2
+ PA2
(pythagoras theorem)
102
= 62
+ PA2
100 = 36 + PA2
PA2
= 100 - 36
= 64
PA = 64 = 8 cm.
O P
A
O
P
10
6
R
Q
P
X
Y
O P
X
Y

Class-X Mathematics
234
EXERCISE - 9.1
1. Fillinthe blanks
(i) A tangent to a circle touches it in................ point (s).
(ii) A line intersecting a circle intwo points is called a.............
(iii) Number oftangents can be drawnto a circle parallelto the given tangent is ......
(iv) The common point ofa tangent to a circle and the circle is called ...............
(v) We candraw ............. tangents to a given circle.
(vi) A circle can have ................ paralleltangents at the most.
2. A tangent PQ at a point P ofa circle ofradius 5 cmmeets a line through thecentre O at
a point Q so that OQ = 13 cm. Find length of PQ.
3. Draw a circle and two lines parallelto a given linedrawn outside the circle such that one
is a tangent and the other, a secant to the circle.
4. Calculate the lengthof tangent from a point 15 cm awayfromthe centre ofa circle of
radius 9 cm.
5. Prove that the tangents to a circle at the end points of a diameter are parallel.
9.3 NUMBER OF TANGENT TO A CIRCLE FROM ANY POINT
To get an idea of the number of tangents from a point on a circle, let us perform the
followingactivity.
ACTIVITY
(i) Draw acircleonapaper. Take apoint
Pinsideit. Canyoudrawatangent to
thecirclethroughthispoint ?Youwill
find that allthe linesthroughthis point
intersect thecircleintwopoints. What
are these ? These are allsecants ofa
circle. So, it is not possible to draw
anytangent to a circle througha point
inside it. (See the adjacent figure)
(ii) Next, take a point P onthe circle and
drawtangentsthroughthispoint. You
have observed that there is only one
tangent to the circle at asuch a point.
(See the adjacent figure)
P
O
(ii)
O
(i)
P

(iii) Now, take a point P outside the circle
and tryto draw tangents to the circle
fromthispoint. What doyou observe?
Youwillfind that youcandrawexactly
two tangents to the circle throughthis
point (See the adjacent figure)
Now, we can summarise these facts as follows :
Case (i) : There is no tangent to a circle passing througha point inside the circle.
Case(ii) : Thereisoneandonlyonetangent toacircleat apointonthecircle.
Case(iii): There are exactlytwo tangents to a circle through a point outside the circle. Inthis
case,AandB are thepointsofcontacts ofthetangents PAandPB respectively.
The length ofthe segment fromthe external point P and the point ofcontact with the
circle is calledthe length of the tangent fromthe point P to the circle.
Note that in the above figure(iii), PAand PBare the lengthofthe tangentsfromPto the
circle. What is the relation between lengths PAand PB?
Theorem-9.2 : The lengthsoftangentsdrawn froman externalpoint to acircle are equal.
Given : Acircle with centre O, Pis a point outside the circle and PAand PB are two tangents
to the circle fromP. (See figure)
To prove : PA = PB
Proof : Join O, A, O, B and O, P.
0
OAP OBP 90
Ð = Ð =
Now inthetwo right triangles
DOAP and DOBP,
OA = OB (radiiofsamecircle)
OP = OP (Common)
Therefore, byR.H.S. Congruencyaxiom,
DOAP @ DOBP
This gives PA= PB (CPCT)
Hence proved.
TRY THIS
Use Pythagoras theoremto write a proof ofthe above theorem.
O
A
P
B
(iii)
A
O
B
P
(Angle between radii and tangents is
900
according to theorem 9.1)

Class-X Mathematics
236
9.3.1. CONSTRUCTION OF TANGENTS TO A CIRCLE FROM AN EXTERNAL POINT
Youhaveseenthatifapointliesoutsidethecircle,therewillbeexactlytwo tangentsto the
circle fromthis point. We shallnow see how to draw these tangents.
Construction : To construct the tangents to a circle fromapoint outside it.
Given : Weare given a circlewith centre ‘O’and a point P outside it. We haveto construct two
tangents fromP to the circle.
Steps of construction :
Step(i) : Join PO and draw a perpendicular
bisector of it. Let M be the midpoint
of PO.
Step (ii) :Taking Mascentre andPMor MO
asradius,drawacircle. Let it intersect
the givencircle at the pointsAand B.
Step (iii) : Join PAand PB. Then PAand PB
are the required two tangents.
Proof:Now, Let usseehow thisconstruction
isjustified.
Join OA. Then PAO
Ð is an angle in
the semicircle and,
therefore, PAO
Ð = 90°.
Can we say that PA OA
^ ?
Since,OAisaradiusofthegivencircle,
PAhas to be a tangent to the circle (By
converse theoremof 9.1)
Similarly, PB is also a tangent to the
circle.
Hence proved.
Some interesting statementsabout tangents and secantsand their proof:
Statement-1:Thecentre ofacircleliesonthebisectoroftheanglebetweentwo tangentsdrawn
froma point outside it. Canyou think how we can prove it?
Proof : Let PQ and PR be two tangents drawn froma point P outsideofthe circle withcentre O
Join O, Q and O, R, triangles OQP and ORP are congruent because,
P
M
O
P
M
O
A
B
P
M
O
A
B

ÐOQP = ÐORP = 90o
(Theorem9.1)
OQ = OR (Radii)
OP iscommon.
This means OPQ OPR
Ð = Ð (CPCT)
Therefore, OP is the bisector angle of QPR
Ð .
Hence, the centre lies on the bisectorofthe angle betweenthe two tangents.
Statement-2 : In two concentric circles, the chord ofthe bigger circle, that touches the smaller
circle isbisected at the point of contact with the smaller circle.
Proof : Consider two concentric circles C1 and C2 with centre O
and a chordAB ofthe larger circle C1, touching the smaller circle
C2 at the point P (See the figure). We need to prove that AP = PB.
Join OP.
ThenAB is a tangent to the circle C2 at PandOPisitsradius.
Therefore, byTheorem9.1,
OP AB
^
Now, DOAPand DOBP are congruent. (Why?) This meansAP= PB. Therefore, OP is
the bisector ofthe chordAB, as the perpendicular from the centrebisects the chord.
Statement-3 : If two tangentsAPandAQare drawn to a circle with centre Ofroman external
pointA, then PAQ 2 OPQ 2 OQP
Ð = Ð = Ð .
Can you see? Why?
Proof : We are givena circle with centre O, an external pointAand two tangentsAPandAQ to
the circle, where P, Q are the points of contact (See figure).
We need to prove that
PAQ 2 OPQ
Ð = Ð
Let PAQ
Ð = q
Now, by Theorem 9.2, AP = AQ.
So DAPQ is an isoscecles triangle
Therefore, ÐAPQ + ÐAQP + ÐPAQ = 180° (Sum of three angles)
Þ
1
APQ AQP (180 )
2
Ð = Ð = °- q
Q
O
R
P
C1
C
2
A
O
P B
A
Q
P
O
q

Class-X Mathematics
238
=
1
90
2
°- q
Also, byTheorem 9.1,
OPA 90
Ð = °
So, OPQ OPA APQ
Ð = Ð - Ð
1 1 1
90 90 PAQ
2 2 2
é ù
= °- - q = q = Ð
ê ú
ë û
This gives
1
OPQ PAQ
2
Ð = Ð .
PAQ 2 OPQ. Similarly PAQ 2 OQP
Ð = Ð Ð = Ð
Statement-4 : If a quadrilateralABCD is drawnto circumscribe a
circle then
AB + CD = AD + BC.
Proof : Since the circle touches the sidesAB, BC, CD and DAof
Quadrilateral ABCD at the points P, Q, R and S respectively as
shown, AB, BC, CD and DA are tangents to the circle.
Since bytheorem9.2, the two tangents to a circle drawnfroma point outside it, are equal,
AP = AS
BP = BQ
DR = DS
and CR = CQ
On adding, we get
AP + BP + DR + CR = AS + BQ + DS + CQ
or (AP + PB) + (CR + DR) = (BQ + QC) + (DS + SA)
or AB + CD = BC + DA.
A
B
C
D
P
Q
S
R

Example-1. Draw a pair oftangents to acircle of radius 5cm which are inclined to each other
at an angle 60°.
Solution : To draw the circle and the two tangents we need to see how we proceed. We only
have the radiusofthe circle andthe angle between the tangents. We do not knowthe distance of
the point fromwhere the tangents are drawnto thecircle and we do not know the lengthofthe
tangentseither.Weknowonlytheanglebetweenthe tangents. Usingthis, we needto find out the
distance ofthepoint outsidethe circlefromwhichwehave to drawthetangents.
To begin, let us consider a circle with centre ‘O’and radius 5cm. Let PAand PB are two
tangents draw from a point ‘P’ outside the circle and the angle between themis 60o
. In this
ÐAPB = 60o
. Join OP.
.
As we know,
OP is the bisector of ÐAPB,
ÐOAP= ÐOPB=
o
60
2
=30o
(Q DOAP @ DOBP)
Now ln DOAP,
sin 30o
=
Opp. side OA
Hyp OP
=
1
2
=
5
OP
(Fromtrigonometric ratio)
OP = 10 cm.
Now we can draw a circle of radius 5 cm with centre ‘O’.
We then mark a point at a distance of 10 cmfrom the centre
ofthecircle. Join OP and complete the constructionas given
in construction 9.2. Hence PAand PB are the required pair
oftangents to the given circle.
Youcanalso trythis constructionwithout using trigonometric ratio.
In OAP; A 90 , P 30 , O 60
D Ð = ° Ð = ° Ð = ° and OA=5 cm. Construct AOP
D to get P.
.
TRY THIS
Draw a pair ofradiiOA and OB in a circle suchthat ÐBOA=120o
. Drawthe bisector
of ÐBOAand draw lines perpendiculars to OAand OB atAand B. These lines meet on
the bisector of ÐBOA at a point whichis the externalpoint and the perpendicular lines are
the required tangents. Construct and Justify.
EXERCISE - 9.2
1. Choosethe correct answer and give justificationfor each.
(i) The angle betweena tangent to a circle and the radius drawn at the point ofcontact is
A
O
B
P
5 cm.
5 cm.
60°
A
O
B
P
10 cm.
M

Class-X Mathematics
240
(a) 60° (b) 30° (c) 45° (d) 90°
(ii) Froma point Q, the lengthof the tangent to a circle is 24 cm. and the distance ofQ from
the centre is 25 cm. The radius ofthe circle is
(a) 7cm (b) 12 cm
(c) 15cm (d) 24.5cm
(iii) IfAPandAQ are the two tangents a circle with centre O so that
POQ 110 ,
Ð = ° then PAQ
Ð is equal to
(a) 60° (b) 70°
(c) 80° (d) 90°
(iv) If tangents PAand PB froma point Pto a circle with centre O are inclined to eachother at
angle of80°, then POA
Ð is equalto
(a) 50° (b) 60°
(c) 70° (d) 80°
(v) In thefigure XYand X1
Y1
aretwo paralleltangents
to a circle with centre O and another tangent AB
with point of contact C intersecting XY at A and
X1
Y1
at B then AOB
Ð =
(a) 80° (b) 100°
(c) 90° (d) 60°
2. Two concentriccirclesofradii5cmand3cmaredrawn. Findthelength
ofthechord ofthelarger circlewhich touchesthesmallercircle.
3. Prove that theparallelogramcircumscribing a circleis a rhombus.
4. AtriangleABC is drawn to circumscribe a circle ofradius 3 cm.
suchthat thesegments BD and DC into whichBC isdivided bythe
point ofcontact D are oflength9 cm. and 3 cm. respectively(See adjacent figure). Find
the sidesAB andAC.
5. Draw a circleofradius 6cm. Froma point 10 cmawayfromits centre, construct the pair
of tangents to the circle and measure their lengths. Verify byusing PythogorasTheorem.
6. Construct atangentto acircleofradius4cmfromapoint ontheconcentriccircleofradius
6cmandmeasure itslength.Also verifythemeasurement byactualcalculation.
7. Draw a circle with the help ofa bangle. Take a point outside the circle. Construct the pair
of tangentsfromthis point to the circle and measure them.Write your conclusion.
A
110°
P
Q
O
O
A
B Y
1
X
1
X Y
C
A
B D
C
3cm.
9 cm.
O

8. InarighttriangleABC,acirclewithasideABasdiameter
is drawn to intersect the hypotenuse AC in P. Prove
that the tangent to the circle at P bisects the side BC.
9. Draw a tangent to a givencircle with center O from a
point ‘R’ outside the circle. How manytangents canbe
drawnto thecircle fromthat point?
Hint :The distance oftwo points to the point ofcontact
is the same.
9.4 SEGMENT OF A CIRCLE FORMED BY A SECANT
We have seen a line and a circle. When a line meets a
circle in only one point, it is a tangent. Asecant is a line which
intersects the circle at two distinct points and the line segment
between the points is a chord.
Here ‘l’is the secant andAB is the chord.
Shankar ismaking apicture bysticking pink
and blue paper. He makes many pictures.
One picture he makes is ofa washbasin. How much paper doeshe need to
make this picture?This picture can be seenin two parts. There is a rectangle
below, but what istheremainingpart?It isthesegment ofacircle.Weknow
how to find the area ofrectangle. How do we find the area ofthe segment?
Inthe followingdiscussion, we willtryto find thisarea.
DO THIS
Shankar made the following picturesalso.
What shapes cantheybe broken into, of which we canfind area easily?
Make some more pictures and think of the shapes they can be divided into different
parts.
A B l
O
A
B Q C
P

Class-X Mathematics
242
Letsusrecalltheformulaeofthe theareaofthefollowinggeometricalfiguresasgiveninthetable.
S.No. Figure Dimensions Area
1. length= l A = lb
breadth = b
2. Side = s A = s2
3. base = b A =
1
2
bh
4. radius = r A = pr2
9.4.1. FINDING THE AREA OF SEGMENT OF A CIRCLE
Swetha made the segments bydrawing secants to thecircle.
As you know, a segment is a region bounded by an arc and a chord. The area that is
shaded ( ) infig.(i) isaminorsegment,a semicircleinfig.(ii) and a majorsegment infig.(iii).
How do we find the area ofa segment? Do the following activity.
Take a circular smallpaper and fold it along a chord and shade the
smaller part asshown in in the figure. What do we callthis smaller part?
It is a minor segment (APB). What do we callthe unshaded portion of
the circle?Obviouslyit is amajor segment (AQB).
You have already come across the sectors and
segments inearlier classes. The portionof some
unshaded part and shaded part (minor segment) is a sector which is the
combinationofatriangle and a segment.
Let OAPB be a sector ofa circle with centreO and radius ‘r’as shown
in the figure. Let the measure of ÐAOB be ‘x’.
O
A B
Q
P
s
s
h
r
b
l
A B
l
A
B
l
A
B
l
(i) (ii) (iii)
b
Q
O
A B
P
r r
x

You know that the area of a circle is pr2
and the angle measured at the centre is 360°.
So,whenthedegreemeasure oftheangleat thecentreis1°, thenareaofthecorresponding
sector is
1
360
°
°
×pr2
.
Therefore, when the degree measure ofthe angle at the centre is x°, the area ofthe sector
is
360
x°
°
×pr2
.
Now let us take the case of the area ofthe segment APB ofa circle with centre ‘O’and
radius ‘r’. You cansee that
Area ofthe segmentAPB =Area of the sector OAPB -Area of DOAB
=
360
x°
°
× pr2
- area of DOAB
TRY THIS
How canyou find thearea ofamajor segment using areaofthe corresponding minor segment?
DO THIS
1. Find the area of sector, whose radiusis 7 cm. withthe given angle:
i. 60° ii. 30° iii. 72° iv. 90° v. 120°
2. The lengthofthe minute handof aclock is14 cm. Find thearea swept bythe minute hand
in10 minutes.
Now, we willsee an example to find area ofsegment of a circle.
Example-1. Find the area ofthe segmentAYB shown in the adjacent figure. It is given that the
radius of the circle is 21 cm and Ð AOB = 1200
(Use p =
22
7
and 3 = 1.732)
Solution : Area of the segment AYB
= Area of sector OAYB -Area of DOAB
Now, area of the sector OAYB =
0
0
120 22
21 21
7
360
´ ´ ´ cm2
= 462 cm2
...(1)
For finding the area of DOAB, draw OM ^ AB as shown inthe figure:-
Note OA= OB. Therefore, byRHS congruence,DAMO @ DBMO.
O
A
B
120°
Y
21 cm.
21 cm.

Class-X Mathematics
244
So, M is the midpoint ofAB and Ð AOM = Ð BOM =
0 0
1
120 60
2
´ =
Let, OM = x cm
So, from DOMA ,
0
OM
cos60 .
OA
=
or,
1
21 2
x
=
0 1
cos60
2
æ ö
=
ç ÷
è ø
Q
or,
21
2
x =
So, OM =
21
2
cm
Also,
AM
OA
= sin 60°
AM 3
=
21 2
0 3
sin 60
2
æ ö
=
ç ÷
è ø
Q
So, AM =
21 3
2
cm.
Therefore AB = 2AM =
2 21 3
2
´
cm. = 21 3 cm
So, Area of DOAB
1
AB OM
2
= ´ ´
1 21
21 3
2 2
= ´ ´ cm2
.
441
3
4
= cm2
. ...(2)
Therefore, area ofthe segmentAYB =
441
462 3
4
æ ö
-
ç ÷
è ø
cm2
.
(Qfrom(1), (2) ]
( ) 2
21
88 21 3 cm
4
= -
= 271.047 cm2
A
B
M
60°
60°
O
21 cm.
21 cm.

Example-2. Find the area of the segments shaded in figure, if PQ = 24 cm., PR = 7 cm. and
QR is the diameter ofthe circle with centre O (Take p=
22
7
)
Solution :Area of the segments shaded =Area ofsector OQPR -Area of triangle PQR.
Since QR is diameter, ÐQPR =90° (Angle ina semicircle)
So, using Pythagoras Theorem
In DQPR, QR2
= PQ2
+ PR2
= 242
+ 72
= 576 + 49
= 625
QR = 625 = 25 cm.
Then radius of the circle =
1
2
QR
=
1
2
(25) =
25
2
cm.
Now, area of semicircle OQPR =
1
2
pr2
=
1 22 25 25
2 7 2 2
´ ´ ´
= 245.53 cm2
..... (1)
Area ofright angled triangle QPR =
1
2
× PR × PQ
=
1
2
× 7 × 24
= 84 cm2
..... (2)
From (1) and (2),
Area ofthe shaded segments = 245.53 - 84
= 161.53 cm2
Example-3. Around table top has sixequaldesigns as shown inthe figure. Ifthe radius ofthe
table top is 14 cm., find the cost ofmaking the designs with paint at the rate of D5 per cm2
.(use
3 = 1.732)
Q
P
R
O
24 cm.
7
c
m
.

Class-X Mathematics
246
Solution :Weknow that the radiusofcircumscribing circle ofa regular hexagon is equalto the
lengthofits side.
Each side of regular hexagon = 14 cm.
Therefore, Area of six design segments = Area of circle - Area of the regular hexagon.
Now, Area of circle = pr2
=
22
7
× 14 × 14 = 616 cm2
..... (1)
Area of regular hexagon = 6 ×
3
4
a2
= 6 ×
3
4
× 14 × 14
= 509.2 cm2
..... (2)
Hence, area of six designs = 616 - 509.21 (from (1), (2)
= 106.79 cm2
.
Therefore, cost of painting the design at the rate of D5 per cm2
= D106.79 × 5
= D533.95
EXERCISE - 9.3
1. In a circle of radius 10 cm, a chord subtends a right angle at the centre. Find the area
of the corresponding: (use p = 3.14)
i. Minor segment ii. Major segment
2. In a circle of radius 12 cm, a chord subtends an angle of 120° at the centre. Find the
area of the corresponding minor segment of the circle (use p = 3.14 and 3 = 1.732)
3. A car has two wipers which do not overlap. Each wiper
has a blade of length 25 cm sweeping through an angle of
115°. Find the total area cleaned at each sweep of the
blades. (use
22
7
p = )
4. Findtheareaoftheshadedregionintheadjacentfigure,where
ABCD is a square of side 10 cm and semicircles are drawn
with each side of the square as diameter (use p = 3.14)
14 cm.
r
A B
C
D
10 cm.

5. Findtheareaoftheshadedregioninfigure, ifABCDisasquare
ofside7cm.andAPDandBPCaresemicircles.(use
22
7
p = )
6. In the figure, OACB is a
quadrant of a circle with centre
O and radius 3.5cm. If
OD=2cm, find the area of the
shaded region.
(use
22
7
p = )
7. AB and CD are respectivelyarcs of two concentric circles of radii
21 cmand 7 cm with centre O (See figure). If ÐAOB = 30°, find
the area ofthe shaded region. (use
22
7
p = )
8. Calculate the area of the designed region in figure, common
between the two quadrants of the circles of radius 10 cm each.
(use p = 3.14)
OPTIONAL EXERCISE
1. Prove that theanglebetweenthe two tangents drawn fromanexternalpoint to a circle is
supplementaryto the angle subtended bythe line - segment joining the points ofcontact
at the centre.
2. PQ is a chord of length 8cm of a circle of radius 5cm.
The tangents at P and Q intersect at a point T (See figure).
Find the length of TP.
3. Prove that opposite sides of a quadrilateral circumscribing a
circlesubtendsupplementaryanglesatthecentreofthecircle.
4. Draw a line segment AB of length 8cm. Taking A as
centre, draw a circle of radius 4cmand taking B as centre,
draw another circle of radius 3cm. Construct tangents to
each circle from the centre of the other circle.
A B
C
D 7 cm.
P
A
C
B
D
O
2 cm
3.5 cm
A B
30°
O
D
C
7
c
m
.
2
1
c
m
.
A B
C
D
10 cm.
10
cm.
P Q
O
T
5 cm.
8 cm.

Class-X Mathematics
248
5. Let ABC be a right triangle in whichAB = 6 cm, BC = 8 cm and B
Ð = 900
. BD is the
perpendicular from B on AC. The circle through B, C, D is drawn. Construct the
tangents fromAto this circle.
6. Find the area of the shaded region in the figure, in which two
circleswithcentresAandB toucheachother at thepointC, where
AC = 8 cm and AB = 3 cm.
7. ABCD is a rectangel
withAB = 14 cm and BC
= 7 cm. Taking DC, BC
and AD as diameters, three semicircles are
drawn as shown in the figure. Find the area of
the shaded region.
In this chapter, we have studied the following points.
1. A tangent to a circle is a line which touches the circle at only one point.
2. The tangent at any point of a circle is perpendicular to the radius through the point of
contact.
3. The lengthsofthe two tangentsfroman externalpoint to acircle are equal.
4. We learnt
a) to construct a tangent to a circle at agiven point when the centre ofthe circle is known.
b) to construct the pair oftangents froman externalpoint to a circle.
5. A secant is a line which intersects the circle at two distinct points and the line segment
between the points is a chord.
6. Area ofsegment ofacircle=Area ofthecorresponding sector -Areaofthe corresponding
triangle.
A
B
C
D
A B
C
14 cm.

10.1 INTRODUCTION
In classes VIII and IX, we have learnt about surface area and volume of regular solid
shapes. We usethemin real life situations to identifywhat we need and what is to be measured
or estimated. For example, to find thequantityofpaint requiredto white washaroom, weneed
the surface areaand not the volume.To find the number ofboxes that would contain a quantity
ofgrain, weneed the volume and not the area.
TRY THIS
1. Consider the following situations. In each find out whether you need to find volume or
area and why?
i. Quantitiyofwater inside a bottle. ii. Canvas needed for making a tent.
iii. Number of bags inside the lorry. iv. Gasfilledin a cylinder.
v. Number ofmatch sticks that canbe put in the match box.
2. State 5 more such examples and ask your friends to choose what they need?
We see so many things of different shapes (combination of two or more regular solids)
aroundus.Housesstandonpillars,storage water tanksarecylindricalandareplacedoncuboidal
foundations, a cricket bat has a cylindricalhandle and a flat main body, etc.Think of different
objects around you. Some of these are shown below:
Mensuration
10

Class-X Mathematics
250
You havelearnt how to findthe surface area andvolume ofsingle regular solids only. We
can however see that other objects canbe seen as combinationsof the solid shapes. So, we now
haveto findtheir surfaceareaandvolumes. Thetableofthesolidshapes, theirareasandvolumes
are givenbelow.
TRY THIS
1. Break the pictures inthe previousfigure into solids ofknownshapes.
2. Think of5 moreobjects aroundyou that canbeseenas a combinationofshapes. Name
the shapes that combined to make them.
Let us recallthe surface areasand volumes ofdifferent solid shapes.
S. Name of Figure Lateral / Curved Total surface Volume Nomen-
No. the solid surface area area clature
1. Cuboid 2h(l+b) 2(lb+bh+hl) lbh l:length
b:breadth
h:height
2. Cube 4a2
6a2
a3
a:side of
the cube
3. Right Perimeter of base Lateral surface area of base -
prism ´ height area+2(area of ´height
the end surface)
4. Regular 2prh 2pr(r+h) pr2
h r:radius of
circular the base
Cylinder h:height
5. Right 1
2 (perimeter of Lateral surfaces 1
3 area of -
pyramid base)´ slant area+area of the base
height the base ´height
6. Right prl pr(l+r) 1
3 pr2
h r:radius of
circular the base
cone h:height
l:slant height
7. Sphere 4pr2
4pr2 4
3 pr3
r:radius
8. Hemisphere 2pr2
3pr2 2
3 pr3
r:radius
a a
a
l
h
b
r
r
r
h
r
l
height
slant
height
h
r

Mensuration 251
Now, let ussee some examples to illustrate the method of finding CSA(Curved Surface
Area), TSA(Total SurfaceArea) ofthe shapes given inthe table.
Example-1. The radius of a conicaltent is 7 meters and its height is 10 meters. Calculate the
length ofcanvas used in making the tent if the width ofcanvas is 2m.
22
Use
7
é ù
p =
ê ú
ë û
Solution : The radius of conical tent is (r) = 7 metres and height (h) = 10 m.
So, the slant height of the cone l2
= r2
+ h2
Þ l = 2 2
r h
+
= 49 100
+
= 149 = 12.2 m.
Now, surface area ofthe tent = prl
2
22
7 12.2 m
7
= ´ ´
= 268.4 m2
.
Area of canvas used = 268.4m2
It is giventhat the width of the canvas = 2m
Length ofcanvas used =
Area 268.4
134.2m
width 2
= =
Example-2. Anoildrumisintheshapeofacylinder havingthefollowingdimensions:diameter
is 2 m. and height is 7 m. The painter charges ` 3 per m2
to paint the drum. Find the total
charges to be paid to the painter for 10 drums ?
Solution : It isgiven that diameter ofthe (oil drum) cylinder= 2 m.
Radius ofcylinder =
2
1 m
2 2
d
= =
Totalsurface area ofa cylindrical drum = 2pr(r + h)
22
2 1(1 7)
7
= ´ ´ +
= 2 ´
22
7
´ 8
7 m
10 m
2 m
7 m

Class-X Mathematics
252
=
352
7
m2
. = 50.28 m2
The totalsurfacearea ofa drum = 50.28 m2
Painting charge per 1m2
= `3.
Cost ofpainting of10 drums = 50.28 ´3´10
= `1508.40
Example-3. A sphere, a cylinder and a cone are ofthe same radius and sameheight. Find the
ratio of their curved surface areas.
Solution : Let r be the common radius ofa sphere, a cone and cylinder.
Height ofsphere = its diameter = 2r.
Then, the height of the cone =height of cylinder =height ofsphere.
= 2r.
The slant height of cone l = 2 2
r h
+
= 2 2
(2 ) 5
r r r
+ =
S1 = Curved surface area of sphere = 4pr2
S2 = Curved surface area of cylinder, 2prh = 2pr ´2r = 4pr2
S3 = Curved surface area of cone = prl = pr ´ 5 r = 5 pr2
Ratio ofcurved surface area is
S1 : S2 : S3 = 4pr2
: 4pr2
: 5 pr2
= 4 : 4 : 5
Example-4. A companywants to manufacture 1000 hemispherical basins from a thin steel
sheet. Iftheradiusofeachbasinis21cm.,findtheareaofsteelsheet required tomanufacturethe
above hemisphericalbasins?
Solution : Radius ofthe hemisphericalbasin (r) = 21 cm
Surface area ofa hemisphericalbasin
= 2pr2
22
2 21
7
= ´ ´ × 21
= 2772 cm2
.

Mensuration 253
Hence, area ofthe steel sheet required for one basin = 2772 cm2
Totalarea ofsteelsheet requiredfor 1000 basins = 2772 × 1000
= 2772000 cm2
= 277.2 m2
Example-5. Aright circular cylinder has base radius 14cmand height 21cm.
Find its (i) area ofbase (or area ofeach end) (ii) curved surface area
(iii) totalsurface areaand (iv) volume.
Solution : Radius ofthe cylinder (r) = 14cm
Height of the cylinder (h) = 21cm
Now (i) Area ofbase(area ofeach end) pr2
=
22
7
(14)2
= 616 cm2
(ii) Curved surface area = 2prh= 2´
22
7
´14´21
= 1848cm2
.
(iii) Totalsurface area = 2´area ofthe base + curved surface area
= 2 ´ 616 + 1848 = 3080 cm2
.
(iv) Volume of cylinder = pr2
h = area ofthe base´height
= 616 ´ 21 = 12936 cm3
.
Example-6. Find the volume and surface area of a sphere of radius 2.1cm (p =
22
7
)
Solution : Radius ofsphere (r) = 2.1 cm
Surface area of sphere = 4pr2
= 2
22 22 21 21
4 (2.1) 4
7 7 10 10
´ ´ = ´ ´ ´
2
1386
55.44 cm
25
= =
Volume of sphere =
3 3
4 4 22
(2.1)
3 3 7
r
p = ´ ´
14 cm
21 cm
2.1 cm

Class-X Mathematics
254
4 22
2.1 2.1 2.1 38.808
3 7
= ´ ´ ´ ´ = cm3
.
Example-7. Find the volume and the total surface area of a hemisphere of radius 3.5 cm.
22
7
æ ö
p =
ç ÷
è ø
Solution : Radius ofsphere (r) is 3.5 cm=
7
2
cm
Volume ofhemisphere =
3
2
3
r
p
3
2 22 7 7 7 539
89.83 cm
3 7 2 2 2 6
= ´ ´ ´ ´ = =
Total surface area = 3pr2
22 7 7
3
7 2 2
= ´ ´ ´ =
231
2
= 115.5 cm2
EXERCISE - 10.1
1. Ajoker’scap is in theformofright circular cone whose base radiusis 7cmand height is 24
cm. Find the area ofthe sheet required to make 10 such caps.
2. Asports companywas ordered to prepare 100 paper cylinders for packing shuttle cocks.
The required dimensions of the cylinder are 35 cm length /height and its radius is 7 cm.
Find the required area ofthick paper sheet needed to make 100 cylinders?
3. Find the volumeofright circular conewith radius 6 cm.and height 7cm.
4. The lateralsurface area ofa cylinder is equalto the curved surface areaofa cone. If their
bases arethe same,find theratio oftheheightofthecylinder to theslant height ofthe cone.
5. Aselfhelp group wants to manufacture joker’s caps of 3cm. radius and 4 cm. height. If
the available paper sheet is 1000 cm2
, then how manycaps can bemanufactured fromthat
paper sheet?
6. A cylinder and cone have bases ofequal radii and are ofequal heights. Show that their
volumes are inthe ratio of3:1.
7. The shape ofsolid iron rod is cylinderical. Its height is 11 cm. and base diameter is 7cm.
Thenfind the totalvolume of50 such rods.
3.5 cm

Mensuration 255
8. A heap ofrice is in theformofa cone ofdiameter 12 m. and height 8 m. Find its volume?
How much canvascloth is required to cover the heap ? (Use p = 3.14)
9. The curved surface area ofa cone is 4070 cm2
and its diameter is70 cm. What isitsslant
height?
10.2 SURFACE AREA OF THE COMBINATION OF SOLIDS
We have seen solids which are made up of combination ofsolids known like
sphere cylinder and cone. We can observe in our real life also like wooden things,
house items, medicine capsules, bottles, oil-tankers etc., We eat ice-creamin our
dailylife. Canyou tell how manysolid figuresare there in it? It is usuallymade up of
cone and hemisphere.
Lets take another example, an oil-tanker /
water-tanker. Is it a single shaped object? You
may guess that it is made up of a cylinder with
two hemisphereat it ends.
If, for some reason you want to find the
surface areas or volumes or capacities of such
objects, howwould you do it?We cannot classify
these shapes under anyofthe solids you have already studied.
As we have seen, the oil-tanker was made up ofa cylinder with two hemispheres stuck at
eitherend.It willlooklikethefollowingfigure:
If weconsider the surface ofthe newlyformed
object, we would be able to see only the curved
surfacesofthetwo hemisphereandthecurvedsurface
ofthe cylinder.
TSAofnewsolid =CSAofonehemisphere + CSAofcylinder +CSAofother hemisphere
HereTSAand CSAstand for ‘totalsurface area’and ‘curved surface area’respectively.
Now let us look at another example.
Devarsha wants to make a toybyputting together a hemisphere and a cone. Let us see the
steps that heshould be going through.

Class-X Mathematics
256
First, he should take a cone and hemisphere ofequalradiiand bring theirflat faces together.
Here, of course, he should take thebase radius ofthe coneequalto the radiusofthe hemisphere,
for the toyto have a smooth surface. So, the steps would be as shown below:
At theend, he got anice round-bottomed toy. Now, ifhewantsto find how muchpaint he
is requiredto colour the surface ofthe toy, forthis he needs to know the totalsurfacearea ofthe
toy, which consists of the CSAofthe hemisphere and the CSAof the cone.
TSA ofthe toy= CSAofhemisphere + CSAofcone
TRY THIS
- Use known solid shapes and make as many objects (bycombining more than two) as
possible that you come across inyour dailylife.
[Hint : Use clay, or balls, pipes, paper cones, boxes like cube, cuboid etc]
THINK AND DISCUSS
Asphere isinscribed in a cylinder. Is the surface ofthesphere equal to
the curved surface ofthe cylinder? Ifyes, explain how?
Example-8. A right triangle, whose base and height are 15 cm. and 20 cm. respectively is
made to revolve about its hypotenuse. Find the volume and surface area ofthe double cone so
formed.
Solution : LetABCbe the right angled triangle suchthat
AB = 15cm and AC = 20 cm
Using Pythagorastheoremin DABCwe have
BC2
= AB2
+ AC2
Step-1 Step-2 Step-3

Mensuration 257
BC2
= 152
+ 202
BC2
= 225 + 400 = 625
BC = 625 = 25 cm.
Let OA= x and OB = y.
In trianglesABO andABC, we have Ð BOA= Ð BAC and Ð ABO = Ð ABC
So , byangle - angle - criterion of similarity, we have DBOA~ DBAC
Therefore,
BO OA BA
BA AC BC
= =
Þ
15
15 20 25
y x
= =
Þ
3
15 20 5
y x
= =
Þ
3
15 5
y
= and
3
20 5
x
=
Þ
3
15
5
y = ´ and
3
20
5
x = ´
Þ y = 9 and x = 12.
Thus, we have
OA = 12 cm and OB = 9cm
Also OC = BC - OB = 25 - 9 = 16 cm
When the ABC is revolved about the hypotenuse. we get a double cone as shown in
figure.
Volume ofthe double cone = volume ofthe cone CAA’+ volume ofthe cone BAA’
2 2
1 1
(OA) OC+ (OA) OB
3 3
= p ´ p ´
2 2
1 1
12 16 12 9
3 3
= p´ ´ + p´ ´
1
144(16 9)
3
= p´ +
1
3.14 144 25
3
= ´ ´ ´ cm3
= 3768 cm3
.
15 cm
. 20 cm.
O
1
5
c
m
.
25 cm.
A
A’
20 cm.
B C
y
x
Note :
2
1
(OA )(OC+OB)
3
p
2
1 22
12 (16 9)
3 7
= ´ ´ ´ +
1 22
144 25
3 7
= ´ ´ ´

Class-X Mathematics
258
Surface area of the doubled cone = (Curved surface area ofcone CAA’)
+ (Curved surface area ofcone BAA’)
= (p´ OAÁC) + (pÓAÁB)
= (p´12´20) + (p´12´15) cm2
= 420 p cm2
2
22
420 cm
7
= ´
= 1320 cm2
.
Example-9. A woodentoyrocket is in theshape ofa conemounted ona cylinderasshown in
the adjacent figure. The height ofthe entire rocket is26 cm, while the height ofthe conicalpart is
6cm. The base of the conical position has a diameter of 5 cm, while the base diameter of the
cylindrical portion is 3 cm. If the conical portion is to be painted orange and the cylindrical
portionis to be painted yellow, findthe area ofthe rocket painted with eachofthese color (Take
p = 3.14)
Solution : Let ‘r’be the radius ofthe base of the cone and its slant height be ‘l’. Further, let r1
be the radius of cylinder and h1 be its height
We have,
r = 2.5 cm., h = 6 cm.
r1 = 1.5 cm. h1 = 20 cm.
Now, 2 2
l r h
= +
Þ 2 2
(2.5) 6
l = +
6.25 36 = 42.25 = 6.5
l = +
Now, area to be painted in orange =
CSA of the cone + base area of the cone -
base areaofthe cylinder
= prl + pr2
- pr1
2
= p{(2.5 × 6.5) + (2.5)2
– (1.5)2
} cm2
= p(20.25) cm2
= 3.14 × 20.25 cm2
= 63.585 cm2
Area to be painted yellow
= Curved surface area ofthe cylinder +Area ofthe base ofthe cylinder
= 2pr1h1 + 2
1
r
p
= pr1 (2h1 + r1)
3 cm
Base of cylinder
Base of cone

Mensuration 259
= 3.14 × 1.5 (2×20+1.5) cm2
= 3.14 ´ 1.5 ´ 41.5 cm2
= 4.71 ´ 41.5 cm2
= 195.465 cm2
.
Therefore, area to be painted yellow = 195.465 cm2
EXERCISE - 10.2
1. Atoyisintheformofaconemountedonahemisphereofthesamediameter.Thediameter
of the base and the height of the cone are 6 cm and 4 cm respectively. Determine the
surface area ofthe toy. [use p = 3.14]
2. A solid is in the formofa right circular cylinder with a hemisphere at one end and a cone at
the other end. The radius ofthe common base is 8cm. and the heights ofthe cylindrical
and conical portions are 10 cm and 6 cm respectivly. Find the total surface area of the
solid. [use p = 3.14]
3. A medicine capsule is in the shape ofa cylinder with two
hemispheres stuck to each of its ends. The length of the
capsule is 14 mm. and the thickness is 5 mm. Find its
surface area.
4. Two cubes eachofvolume 64 cm3
are joined end to end together. Find the surface area
oftheresulting cuboid.
5. Astoragetank consistsofacircular cylinder witha hemisphere stuck oneither end. Ifthe
externaldiameterofthe cylinder be1.4 m. and itslengthbe 8 m.find the cost ofpainting it
on the outside at rate of `20 per m2
.
6. A sphere, a cylinder and a cone have the same radius and same height. Find the ratio of
theirvolumes.
[Hint : Diameter ofthe sphere is equalto theheights ofthe cylinder and the cone.]
7. Ahemisphere iscut out fromoneface ofa cubicalwoodenblock suchthat
the diameter ofthe hemisphere is equalto the side ofthe cube. Determine
the totalsurface area ofthe remaining solid.
8. A wooden articlewas made byscooping out a hemiphere fromeach end of
a solid cylinder, as showninthefigure. Iftheheight ofthecylinderis 10cm.
and its radius of the base is of 3.5 cm, find the total surface area of the
article.

Class-X Mathematics
260
10.3 VOLUME OF COMBINATION OF SOLIDS
Let us understand volume of a combined solid
throughanexample.
Suresh runs an industry in a shed which is in the
shape ofa cuboid surmounted bya halfcylinder. The base
ofthe shed isofdimensions 7 m. ´15 m.and theheight of
the cuboidalportionis 8 m. Find the volume ofair that the
shedcanhold? Furthersupposethemachineryintheshed
occupies a totalspace of300m3
and thereare 20workers,
each of whom occupies about 0.08 m3
space on an
average. Thenhow much air is inthe shed ?
The volume ofair inside the shed (when there are
neither people nor machinery) is given bythe volume of
air inside the cuboid and inside the half cylinder taken
together.The length, breadthand height ofthe cuboid are
15 m., 7 m. and 8 m. respectively. Also the diameter of the halfcylinder is 7 m. and its height is
15 m.
So the required volume= volume ofthe cuboid +
1
2
volume ofthe cylinder.
.
3
1 22 7 7
15 7 8 15 m
2 7 2 2
é ù
= ´ ´ + ´ ´ ´ ´
ê ú
ë û
= 1128.75m3
.
Next, the toalspace occupied bythe machinery
= 300 m3
.
And thetotalspace occupied bythe workers
= 20´0.08 m3
= 1.6m3
Therefore, the volume of the air, inside the shed when there are machineryand workers
= 1128.75 - (300.00 + 1.60)
= 1128.75 - 301.60 = 827.15 m3
Note : Incalculating the surface area ofcombination of solids, we can not addthe surface areas
of the two solids because some part of the surface areas disappears in the process of joining
them. However, this willnot be the case whenwe calculate the volume. The volume ofthe solid
formed byjoiningtwo basic solids willacutallybe the sumofthe volumes ofthe constituents as
seen inthe example above.

Mensuration 261
TRY THIS
1. Ifthe diameter ofthe cross - section ofa wire is decreased by5%, bywhat percentage
should the length be increased so that the volume remainsthe same ?
2. Surface areas ofa sphere and cube are equal. Thenfind the ratio oftheir volumes.
Let us see some more examples.
Example-10. Asolid toyis inthe formofaright circular cylinder withhemisphericalshape at
one end and a cone at the other end. Their common diameter is 4.2 cm and the height of the
cylindricalandconicalportionsare 12cmand 7cmrespectively. Find the volume ofthe solid toy.
22
Use
7
æ ö
p =
ç ÷
è ø
.
Solution : Let height ofthe conicalportion h1 = 7cm
The height ofcylindricalportion h2 = 12 cm
Radius (r) =
4.2
2
= 2.1 =
21
10
cm
Volume ofthe solid toy
= Volume ofthe Cone + Volume of the Cylinder + Volume of the Hemisphere.
=
2 2 3
1 2
1 2
3 3
r h r h r
p + p + p
=
2
1 2
1 2
3 3
r h h r
é ù
p + +
ê ú
ë û
2
22 21 1 2 21
7 12
7 10 3 3 10
æ ö é ù
= ´ ´ ´ + + ´
ç ÷ ê ú
è ø ë û
22 441 7 12 7
7 100 3 1 5
é ù
= ´ ´ + +
ê ú
ë û
22 441 35 180 21
7 100 15
+ +
é ù
= ´ ´ ê ú
ë û
3
22 441 236 27258
218.064 cm .
7 100 15 125
= ´ ´ = =
h1
h2
r
cm
cm
cm

Class-X Mathematics
262
Example-11. A cylindrical container is filled with ice-cream whose diameter is 12 cm and
height is15 cm. The wholeice creamisdistributed to 10 children by filling in equalcones
andforminghemisphericaltops. Ifthe height oftheconicalportionistwicethe diameterof
its base, find the diameter of the ice cream cone.
Solution : Let the radius ofthe base ofconicalice cream = x cm
diameter = 2x cm
Then, the height of the conicalice cream
= 2 (diameter) = 2(2x) = 4x cm
Volume of ice creamcone
= Volumeofconicalportion +Volume ofhemisphericalportion
=
1
3
pr2
h +
2
3
pr3
2 3
1 2
(4 )
3 3
x x x
= p + p
3 3 3
4 2 6
3 3
x x x
p + p p
= =
= 2px3
cm3
Diameter ofcylindricalcontainer =12 cm
Its height (h) = 15 cm
Volume ofcylindricalcontainer = pr2
h
= p(6)2
15
= 540p cm3
Number of childrento whomice creamis given = 10
Volumeof cylindricalcontainer
Volume of one icecreamcone
= 10
Þ 3
540
2 x
p
p
= 10
2px3
× 10 = 540p
Þ
3 540
27
2 10
x = =
´
x cm
x cm

Mensuration 263
Þ x3
= 27
Þ x3
= 33
Þ x = 3
Diameter of ice cream cone 2x = 2(3) = 6cm
Example-12. A solid consisting of a right circular cone standing on a hemisphere, is placed
upright inaright circular cylinder fullofwaterandtouchingthe bottom. Findthe volumeof
water left in the cylinder, giventhat the radius ofthe cylinder is 3 cm. and its height is 6cm.
The radius ofthe hemisphere is2 cm. and the height ofthe cone is 4 cm.
22
Take
7
æ ö
p =
ç ÷
è ø
.
Solution : In the figure drawn here,
ABCD is a cylinder, LMN is a hemisphere and OLM is a cone
We know that when a solid is immersed inthe water, thenwater
displaced equalto the volume ofthe solid.
Volume of the cylinder = pr2
h = p ´ 32
´ 6 = 54 p cm3
Volume of the hemisphere =
3 3 3
2 2 16
2 cm
3 3 3
r
p = ´p´ = p
Volume ofthe cone =
2 2 3
1 1 16
2 4 cm
3 3 3
r h
p = ´ p´ ´ = p
Volume ofthesolid figure =
16 16
π + π
3 3
=
32
π
3
Volume ofwaterleft in the cylinder
= Volume ofCylinder -Volume ofsolid figure immersed
= 54p -
32
3
p
=
162 32 130
3 3
p - p p
=
A B
C
D
O
N
L M
3 3
2
2 2
4

Class-X Mathematics
264
3
130 22 2860
136.19cm
3 7 21
= ´ = =
Example-13. A cylindricalpencilis sharpened to produce a perfect cone at oneend with no
over allloss ofits length. The diameter ofthe pencilis 1cmand the length ofthe conical
portion is2cm. Calculate the volume of the peels. Give your answer correct to two places
ifit isin decimal
355
use
113
é ù
p =
ê ú
ë û
.
Solution : Diameter ofthe pencil= 1cm
So, radius of the pencil(r) = 0.5 cm
Length ofthe conicalportion = h= 2cm
Volume ofpeels = Volume ofcylinder oflength 2 cmand base radius 0.5 cm.
- volume of the cone formed by this cylinder
=
2 2 2
1 2
3 3
r h r h r h
p - p = p
=
2 3 3
2 355
(0.5) 2 cm 1.05 cm
3 113
´ ´ ´ =
EXERCISE-10.3
1. An ironpillar consists of acylindricalportionof 2.8 m height and 20 cmindiameter and
a cone of42 cm height surmounting it. Findtheweight ofthepillarif1cm3
ofiron weighs
7.5 g.
2. A toyismade in the formofhemisphere surmounted bya right cone whose circular base is
joined withthe plane surface ofthe hemisphere. Theradius ofthe baseof the cone is7 cm
and its volume is
3
2
ofthe hemisphere. Calculate the height ofthe cone and the surface
area of the toy correct to 2 places ofdecimal
1
Take 3
7
æ ö
p =
ç ÷
è ø
.
3. Find the volume ofthe largest right circular cone that canbecut out ofa cube whose edge
is 7 cm.
1 .
cm
2
.
c
m

Mensuration 265
4. Acylindericalmugof radius 5cmand height 9.8 cm
is fullofwater. Asolid in the formofright circular
cone mountedona hemisphere isimmersed into the
mug. The radius of the hemisphere is 3.5 cm and
height ofconicalpart5cm. Findthevolumeofwater
left in the tub
22
Take
7
æ ö
p =
ç ÷
è ø
.
5. Intheadjacentfigure,theheight ofasolidcylinderis
10cmanddiameter is7cm. Twoequalconicalholes
ofradius 3cmand height 4 cmare cut offas shown
the figure. Find the volume ofthe remaining solid.
6. Spherical marbles ofdiameter 1.4 cm are dropped into a cylindricalbeaker ofdiameter 7
cm, whichcontainssome water. Find the number ofmarblesthat should bedropped into
the beaker, so that water levelrises by 5.6 cm.
7. Apenstandis made ofwoodinthe shape ofcuboid withthree conicaldepressions to hold
the pens. The dimensions ofthe cuboid are 15cmby10 cmby3.5 cm. Theradius of each
of the depressionis 0.5 cmand the depthis1.4cm. Find thevolume of wood inthe entire
stand.
10.4 CONVERSION OF SOLID FROM ONE SHAPE TO
ANOTHER
Awomenselfhelp group (DWACRA) preparescandles bymelting
down cuboid shapewax. In gunfactories sphericalbullets are made
bymelting solidcubeoflead, goldsmithprepares variousornaments
by melting cubiod gold biscuts. In all these cases, the shapes of
solids areconvertedinto anothershapes.Inthisprocess, thevolume
always remainsthe same.
How does this happen? If you want a candle of any specialshape, you have to heat thewax in
metalcontainer tillit iscompletelymelted. Thenyoupourit into another containerwhichhasthe
specialshape that you wanted.
3 cm.
4 cm.
4 cm.
3 cm.

Class-X Mathematics
266
For example, letsus take a candleinthe shape ofsolid cylinder, melt it andpour whole of
the moltenwaxinto another container shapedlike a sphere. Oncooling, you willobtaina candle
in the shape ofthe sphere. The volume ofthenew candle willbe the same as the volume ofthe
earlier candle. This is what we have to remember when we come across objects which are
converted fromone shape to another, or whena typeofliquid whichoriginallyfilled a container
ofa particular shape ispoured into another container ofadifferent shape or sizeasyou observe
inthefollowingfigures.
THINK AND DISCUSS
Which barrel shown in the adjacent figure
canhold more water?Discuss withyour friends.
To understand what has been discussed, let us consider some examples.
Example-14. Acone ofheight 24cm and radius of base 6cmis made up of modelling clay. A
child moulds it in the formofa sphere. Find the radius ofthe sphere.
Solution : Volume ofcone =
1
6 6 24cm
3
3
´p´ ´ ´
If r is the radius ofthe sphere, then itsvolume is
3
4
3
r
p
Sincethevolumeofclayintheformoftheconeandthesphereremainsthesame,wehave
3
4
3
r
p =
1
π
3
× 6 × 6 × 24
r3
= 3 ´ 3 ´ 24 = 3 × 3 × 3 × 8
r3
= 33
´ 23
r = 3 ´ 2 = 6
Therefore the radius of the sphere is 6cm.
4
1
4
1

Mensuration 267
DO THIS
1. A copper rod ofdiameter 1 cm. and length8 cm. is drawn into a wire oflength18mof
uniformthickness. Find the thicknessof the wire.
2. Pravali’s house has a water tank in the shape ofa cylinder on the roof. This is filled by
pumpingwaterfromasump(anundergroundtank)whichisintheshapeofacuboid. The
sump hasdimensions1.57 m. ´1.44m. ´ 1.5 m. The water tank has radius 60 cm. and
height 95cm. Find theheight ofthe waterleft in the sump after the water tankhas been
completelyfilled with water fromthe sump whichhad been fullofwater. Compare the
capacityofthe tank withthat ofthe sump. (p =3.14)
Example-15. Thediameter ofthe internaland externalsurfaces ofa hollow hemisphericalshell
are 6cm. and 10 cm. respectively. It is melted and recast into asolid cylinder ofdiameter 14 cm.
Find the height ofthe cylinder.
Solution : Outer radius of hollow hemispherical shell R =
10
2
= 5 cm.
Internalradius ofhollow hemisphericalshell (r) =
6
2
= 3cm.
Volume ofhollow hemisphericalshell
= Externalvolume- Internalvolume
3 3
2 2
R
3 3
r
= p - p
3 3
2
(R )
3
r
= p -
3 3
2
(5 3 )
3
= p -
2
(125 27)
3
= p -
3 3
2 196
98cm cm
3 3
p
= p´ = ...(1)
Since, this hollow hemispherical shell is melted and recast into a solid cylinder. So
their volumes must be equal
Diameter of cylinder = 14 cm. (Given)
So, radius of cylinder = 7 cm.
r
r
10 cm. 6 cm.

Class-X Mathematics
268
Let the height ofcylinder = h
volume of cylinder = pr2
h
= p ´ 7 ´ 7 ´ h cm3
= 49ph cm3
...(2)
According to givencondition
volume ofhollowhemisphericalshell= volumeofsolid cylinder
196
49
3
h
p = p [Fromequatiion(1) and (2)]
Þ
196 4
cm.
3 49 3
h = =
´
Hence, height ofthe cylinder = 1.33 cm.
Example-16. Ahemisphericalbowlofinternalradius 15 cmcontains a liquid. The liquid is to
be filled into cylindrical bottles of diameter 5 cm and height 6 cm. How many bottles are
necessary to emptythe bowl?
Solution : Volume ofhemisphere =
3
2
3
r
p
Internalradius ofhemisphere r = 15 cm
Volume ofliquidcontained in hemisphericalbowl
=
3 3
2
(15) cm
3
p
= 2250 p cm3
.
This liquidis to be filledincylindricalbottles andthe height ofeachbottle (h) =6 cm.
and radius (R) =
5
2
cm
Volume of 1cylindrical bottle = pR2
h
2
5
6
2
æ ö
= p´ ´
ç ÷
è ø
3 3
25 75
π 6cm π cm
4 2
= ´ ´ = .

Mensuration 269
Number ofcylindrical bottles required =
Volume of hemispherical bowl
Volume of 1 cylindrical bottle
=
2250
75
2
p
p
=
2 2250
60
75
´
= .
Example-17. The diameter ofa metallic sphere is 6cm. It is melted and drawn into a long wire
having a circular cross section ofdiameter as 0.2 cm. Find the lengthofthe wire.
Solution : Diameter ofmetallic sphere = 6cm
Radius ofmetallic sphere = 3cm
Now, diameter of cross – sectionofcylindricalwire = 0.2 cm.
Radius of cross section ofcylinder wire = 0.1 cm.
Let the length ofwire be lcm.
Sincethe metallicsphereis converted into a cylidricalshaped wire oflength hcm.
Volume ofthe metalused inwire = Volume ofthe sphere
2 3
4
(0.1) 3
3
h
p´ ´ = ´p´
2
1 4
27
10 3
h
æ ö
p´ ´ = ´p´
ç ÷
è ø
1
π 36π
100
h
´ ´ =
36 100
h
p´
=
p
cm
= 3600 cm = 36 m
Therefore, the length of the wireis 36 m.
6 cm.
0.2 cm.

Class-X Mathematics
270
Example-18. How many spherical balls can be made out ofa solid cube oflead whose edge
measures 44 cmand each ballbeing 4 cm. in diameter.
Solution : Side of lead cube = 44 cm.
Radius of sphericalball =
4
2
cm. = 2 cm.
Now volume of a spherical ball=
3
4
3
r
p
=
3 3
4 22
2 cm
3 7
´ ´
3
4 22
8 cm
3 7
= ´ ´
Let the number ofballs be ‘x’.
Volume of x sphericalball
3
4 22
8 cm
3 7
x
= ´ ´ ´
It is clear that volume of x sphericalballs = Volume oflead cube
3
4 22
8 = (44)
3 7
x
Þ ´ ´ ´
4 22
8 = 44 44 44
3 7
x
Þ ´ ´ ´ ´ ´
44 44 44 3 7
=
4 22 8
x
´ ´ ´ ´
Þ
´ ´
x = 2541
Hence, totalnumber ofsphericalballs = 2541.
Example-19. Awomen selfhelp group (DWACRA) is supplied a rectangular solid (cuboid
shape) ofwax block with dimensions 66 cm, 42 cm, 21 cm, to preparecylindricalcandles each
4.2 cmindiameter and 2.8 cmofheight. Find the number ofcandles prepared using this solid.
Solution : Volumeofwax inthe rectangular solid = lbh
= (66 ´42´21) cm3
.
Radius ofcylindricalcandle =
4.2
2
cm. = 2.1 cm.
Height ofcylindricalcandle = 2.8 cm.

Mensuration 271
Volume ofcandle = pr2
h
=
2
22
(2.1) 2.8
7
´ ´
Let x be the number ofcandles
Volume ofx cylindricalwaxcandles =
22
2.1 2.1 2.8
7
x
´ ´ ´ ´
Q Volume of x cylindricalcandles = volume ofwaxin rectangular shape
22
2.1 2.1 2.8 66 42 21
7
x
´ ´ ´ ´ = ´ ´
66 42 21 7
22 2.1 2.1 2.8
x
´ ´ ´
=
´ ´ ´
= 1500
Hence, the number of cylindricalwaxcandles that can be prepared is 1500.
EXERCISE - 10.4
1. A metallic sphere of radius 4.2 cm is melted and recast into the shape of a cylinder of
radius6cm. Find the height ofthe cylinder.
2. Three metallic spheres of radii6 cm, 8 cmand 10 cmrespectivelyare melted together to
forma single solid sphere. Find the radius ofthe resulting sphere.
3. A 20mdeep wellof diameter 7 mis dug and the earth got bydigging is evenlyspread out
to forma rectangular platformof base 22 m ´ 14 m. Find the height ofthe platform.
4. A well of diameter 14 m is dug 15 m deep. The earth taken out of it has been spread
evenlyto formcircular embankment allaroundthewallofwidth7m. Findtheheight ofthe
embankment.
5. Acontainer shaped a right circular cylinder having diameter 12 cmand height 15 cmis full
of ice cream. The icecream is to be filled into cones of height 12 cm and diameter
6 cm, making a hemispherical shape onthe top. Find the number of suchcones which can
be filled with ice cream.
6. How manysilver coins, 1.75 cm in diameter and thickness 2 mm, need to be melted to
form a cuboid ofdimensions 5.5 cm ´10 cm ´ 3.5 cm?
7. A vesselis inthe formofaninverted cone. Itsheight is 8 cm. and the radius ofits top is 5
cm. It is filled with water up to the rim. When lead shots, each of which is a sphere of
radius 0.5cmare dropped into the vessel,
1
4
ofthe water flowsout. Find the number of
lead shots dropped into the vessel.
8. A solid metallic sphere of diameter 28 cm is melted and recast into a number of
smaller cones, each of diameter 4
2
3
cmand height 3cm. Find the number ofcones so
formed.

Class-X Mathematics
272
OPTIONAL EXERCISE
1. A golf ballhas diameter equal to 4.1 cm. Its surface has 150 dimples each ofradius 2 mm.
Calculatetotalsurfaceareawhichisexposedto thesurroundings.(Assumethat thedimples
are allhemispherical)
22
7
p
é ù
=
ê ú
ë û
2. A cylinderofradius12 cmcontains water to adepth of20 cm. When a sphericaliron ball
is dropped in to the cylinder,the levelofwaterisraised by6.75 cm. Find the radiusofthe
ball.
22
7
p
é ù
=
ê ú
ë û
3. A solid toyis inthe formof a right circular cylinder with a hemispherical shapeat one end
andaconeat theotherend.Their commondiameteris4.2 cm. andheightsofthecylindrical
and concial portion are 12 cm. and 7 cm. respectively. Find the volume of the solid
toy.
22
7
p
é ù
=
ê ú
ë û
4. Three metalcubes with edges 15 cm., 12 cm. and 9 cm. respectivelyare melted together
and formed into a single cube. Find the diagonalofthis cube.
5. Ahemisphericalbowlofinternaldiameter36 cm. containsaliquid. Thisliquid isto be filled
in cylindricalbottles of radius 3 cm. and height 6 cm. How manybottles are required to
emptythe bowl?
WHAT WE HAVE DISCUSSED.
1. The volume of the solid formed by joining two or more basic solids is the sum of the
volumes ofthe constituents.
2. In calculating the surface area ofa combination of solids, we can not add the surface area
ofthe two constituents, because some part ofthe surface area disappearsonjoining them.
Suggested Projects
Make an open box from a 20cm by 20cm piece of cardboard by cutting out four squares
fromcorners folding the flaps. What is the biggest volume ofbox you can makein this way?
Can you find a relation between the size of paper and the size of the square cutout that
producesthemaximumvolume.
Extension: You canextend this bytaking a rectangular sheet of paper instead ofa square
sheet of paper.

11.1 INTRODUCTION
We have learnt about triangles and their properties in previous
classes. There, weobserveddifferent dailylifesituationswheretriangles
are used.
Let’sagain look at some ofthe daily life examples.
l Electricpolesarepresent everywhere.Theyareusuallyerected
byusing a metalwire. The pole, wire and the ground form a
triangle. But, ifthe lengthofthe wire decreases,what willbe
the shape ofthe triangle and what willbe the angleofthe wire
withthe ground ?
l Apersoniswhitewashingawallwiththehelp ofaladderwhich
is kept as shown in theadjacent figure on
left. Ifthe person wants to paint at a higher position, what willthe
persondo? What willbe the change in angle oftheladder with the
ground ?
l In the temple at Jainath inAdilabad district, which was built in
13th
century, the first rays ofthe Sun fall at the feet of the Idol of
SuryanarayanaSwamiinthemonthofDecember.Thereisarelation
between distance of Idolfrom the door, height of the hole on the
door fromwhichSunrays are entering and angle ofsunrays inthat
month. Isthere anytriangle forminginthiscontext?
l In a play ground, children like
to slideon a slider and slider is
on a defined angle from earth.
What willhappento thesliderif
we change the angle? Will
children stillbe able to playon
it?
q
q
q
Trigonometry
11

Class-X Mathematics
274
Theaboveexamplesaregeometricallyshowingtheapplicationpart oftrianglesinourdaily
life andwe can measure the heights, distances and slopes by using the properties of triangles.
These types ofproblems are part of‘trigonometry’whichis a branchofmathematics.
Now look at the example of a person who is white washing the wall with the help of a
ladder asshownin the previousfigure. Let us observethe following conditions.
We denote the foot ofthe ladder byAand top ofit byC and the point ofintersection of
the walland line throughbaseofthe ladder asB. Therefore, DABC isa right angledtrianglewith
right angle at B. Let the angle between ladder and base be q.
1. If the person wants to white
wash at a higher point on the
wall-
l What happens to the angle
made bythe ladder with the
ground?
l What will be the change in
the distanceAB?
2. If the person wants to white
wash at a lower point on the
wall-
l What happens to the angle made bythe ladder withthe ground?
l What willbe the change inthe distanceAB?
We have observed in the above example of a personwho was white washing.When he
wants to paint at higher or lower points, he should change the position ofladder. So, when ‘q’is
increased, the height also increases and the base decreases. But, when qis decreased,theheight
also decreases and the base increases. Do you agree withthis statement?
Here, we have seen a right angle triangleABC now let’s name the sides again because
trigonometric ratios ofanglesare based on sides only.
q
A B
C

Trigonometry 275
11.1.1 NAMING THE SIDES IN A RIGHT TRIANGLE
Let’s take a right triangleABC as show in the figure.
In triangleABC, we can consider Ð CAB asAwhere angleAis an acute angle. Since
AC is the longest side, it is called “hypotenuse”.
Here you observe the position
ofside BC withrespect to angleA. It is
opposite to angleAand we cancallit as
“opposite side of angle A”. And the
remaining side AB can be called as
“Adjacent side of angleA”
AC = Hypotenuse
BC = Opposite side of angleA
AB =Adjacent side ofangleA
DO THIS
Identify“Hypotenuse”, “Opposite side” and “Adjacent side” for the given angles in
the given triangles.
1. For angle R 2. (i) For angle X
(ii) For angleY
TRY THIS
Write lengths of“Hypotenuse”, “Opposite side” and
“Adjacent side” for the given angles inthe given triangles.
1. For angle C
2. For angleA
What do youobserve? Is thereanyrelation between the opposite side of theangleAand
adjacent side ofangle C? Like this,suppose you are erecting a pole bygiving support of strong
ropes. Is thereanyrelationship between thelength of the rope and the length ofthe pole? Here,
we haveto understand the relationshipbetweenthe sides and angles we will studythis under the
sectioncalledtrigonometric ratios.
P
R
Q
Z
X Y
C
A
B
C
A B
q

Class-X Mathematics
276
11.2 TRIGONOMETRIC RATIOS
Wehaveseensomeexamplesinthebeginningofthechapterwhicharerelatedto our daily
life situations.Let’sknow about thetrigonometricratiosand howtheyaredefined.
ACTIVITY
1. Draw a horizontalline ona paper.
2. Let the initialpoint beAand mark other points B, C, D and E at a distance
of3cm, 6cm, 9cm, 12 cmrespectively fromA.
3. Draw the perpendiculars BP, CQ, DR and ES of
lengths 4cm, 8cm, 12cm, 16cm from the
points B, C, D and E respectively.
4. Then join AP, PQ, QR and
RS.
5. Find lengths ofAP, AQ, AR and AS.
Name of Name of Length of Length of Length of Opposite side Adjacent side
triangle the triangle hypotenuse opposite side adjacent side Hypotenuse Hypotenuse
Then find the ratios of
BP CQ DR
, ,
AP AQ AR
and
ES
AS
.
Did you get the same ratio as
4
5
?
Similarly try to find the ratios
AB AC AD
, ,
AP AQ AR
and
AE
AS
? What do you observe?
ABP
D
ACQ
D
ADR
D
AES
D
S
A
B C D E
P
Q
R
q
Y
X

Trigonometry 277
11.2.1 DEFINING TRIGONOMETRIC RATIOS
In the above activity, when we observe right angle trianglesABP,ACQ,ADR andAES,
Ð A is common, Ð B, Ð C, Ð D and Ð E are right angles and Ð P,
, Ð Q, Ð R and Ð S are
also equal. Hence, we cansay that trianglesABP,ACQ,ADR andAES are similar triangles.
When we observe the ratio of opposite side of angleAand hypotenuse in a right angle triangle
and theratio ofsimilar sides in another triangle, it is found to be constant in allthe above right
angle trianglesABP,ACQ,ADR andAES. The ratios
BP CQ DR
, ,
AP AQ AR
and
ES
AS
are named as
“sineA” or simply “sinA” in those triangles. Ifthe value ofangleAis ‘x’, then the ratio would
be “sin x”.
Hence, we can conclude that the ratio of opposite side of an angle (measure of the angle)
and length of the hypotenuse is constant in all similar right angle triangles. This ratio will
be named as “sine” of that angle.
Similarly, when we observe the ratios
AB AC AD
, ,
AP AQ AR
and
AE
AS
, it is also found to be
constant. And these are the ratios of the adjacent sides ofthe angleAand hypotenuses in right
angle triangles ABP, ACQ, ADR and AES. So, the ratios
AB AC AD
, ,
AP AQ AR
and
AE
AS
will be
named as “cosineA” orsimply “cosA” inthose triangles. Ifthe value oftheangleAis “x”,then
the ratio would be “cos x”
Hence, we can also conclude that the ratio of the adjacent side of an angle (measure of the
angle) and length of the hypotenuse is constant in all similar right triangles. This ratio
will be named as “cosine” of that angle.
Similarly, the ratio of opposite side and adjacent side of an angle is constant and it can be
named as “tangent” of that angle.
LET’S DEFINE RATIOS IN A RIGHT ANGLE TRIANGLE
Consideraright angletriangleABChavingrightangle atB asshowninthe followingfigure.
Then, trigonometric ratios ofthe angleAin right angle triangleABC are defined as follows :

Class-X Mathematics
278
sine of Ð A = sinA =
Length of the sideoppositetoangleA BC
Lengthof hypotenuse AC
=
cosine of Ð A= cosA=
Length of the sideadjacent toangleA AB
Length of hypotenuse AC
=
tangent of Ð A= tanA=
Length of the sideoppositetoangleA BC
Length of the side adjacent toangle A AB
=
DO THIS
1. Find (i) sin C (ii) cos C and
(iii) tanC in the adjacent triangle.
2. Inatriangle XYZ, ÐY isright angle,
XZ = 17 m and YZ = 15 cm, then find
(i) sin X (ii) cos Z (iii) tan X
3. In a triangle PQR with right angle at Q, the value of ÐP is x, PQ = 7 cmand QR =
24 cm, then find sin x and cos x.
TRY THIS
In a right angle triangle ABC, right angle is at C. BC + CA = 23 cm and
BC - CA= 7cm, then find sinAand tan B.
THINK AND DISCUSS
Discuss among your friends
(i) Does sin x =
4
3
exist for some value of angle x?
(ii) The value ofsinAand cosAis always less than 1. Why?
(iii) tanAis product of tan andA.
There are three more ratios defined in trigonometry which are considered as
multiplicative inverses of the above three ratios.
C
A B
C
A
B

Trigonometry 279
Multiplicativeinverseof “sineA”is “cosecantA”, simplywritten as“cosecA”, it is also
sometimes written as cscA
i.e., cosec A=
1
sin A
Similarly, multiplicativeinverseof “cosA” is secantA” (simplywrittenas“secA”) and
that of “tanA” is “cotangentA(simplywrittenas cotA)
i.e., sec A =
1
cos A
and cot A =
1
tan A
How can you define ‘cosec’ in terms of sides?
If sin A=
Opposite sideof theangleA
Hypotenuse
,
then cosec A =
Hypotenuse
Opposite sideof theangleA
TRY THIS
Express secAand cosAin terms of sides of right angle triangle.
THINK AND DISCUSS
l Is sin A
cosA
equal to tanA? l Is cosA
sin A
equal to cot A?
Let us see some examples
Example-1. If tanA=
3
4
, thenfind the other trigonometric ratio ofangleA.
Solution : Given tanA=
3
4
Hence tanA=
Opposite side
Adjacent side
=
3
4
Therefore, opposite side : adjacent side = 3:4
For angleA, opposite side = BC = 3k
Adjacent side =AB = 4k (where k is any positive number)
Now, wehave intriangleABC(byPythagoras theorem)
C
A B

Class-X Mathematics
280
AC2
= AB2
+ BC2
= (3k)2
+ (4k)2
= 25k2
AC = 2
25k
= 5k = Hypotenuse
Now, we caneasilywrite the otherratiosoftrigonometry
sin A =
3 3
5 5
=
k
k
, cos A=
4 4
5 5
=
k
k
Hence cosec A =
1 5
sin A 3
= , sec A =
1 5
,
cosA 4
= cot A=
1 4
tan A 3
= .
Example-2. If ÐAand ÐPare acute anglessuch that sinA= sin P then prove that ÐA=
=ÐP
Solution : Given sinA= sin P
we have sin A=
BC
AC
and sin P =
QR
PQ
Then
BC QR
AC PQ
=
Let,
BC QR
AC PQ
= = k .....(1)
ByusingPythagorastheorem
( )
2 2
2 2 2 2 2
2 2 2 2 2
AC 1
AB AC BC AC AC
PR PQ QR PQ PQ
k
k
k
-
- -
= = =
- - ( )
2 2
PQ 1 k
-
AC
PQ
=
(From (1))
Hence, AC AB BC
PQ PR QR
= = then DABC : DPQR
Therefore, A P
Ð = Ð
Example-3. Consider a triangle PQR, right angled at R, in which PQ = 29 units, QR = 21
units and Ð PQR = q, then find the values of
C
A B
Q
P R

Trigonometry 281
(i) cos2
q + sin2
q and (ii) cos2
q - sin2
q
Solution : ln PQR, we have
2 2 2 2
PR PQ QR (29) (21)
= - = -
= 400 = 20 units
sinq =
PR 20
PQ 29
=
cos q =
QR 21
PQ 29
=
Now (i) cos2
q + sin2
q =
2 2
20 21 441 400
1
29 29 841
+
æ ö æ ö
+ = =
ç ÷ ç ÷
è ø è ø
(ii) cos2
q - sin2
q =
2 2
20 21 41
29 29 841
-
æ ö æ ö
- =
ç ÷ ç ÷
è ø è ø
EXERCISE - 11.1
1. In right angle triangleABC, 8 cm, 15 cmand 17 cmare the lengths ofAB, BC and CA
respectively. Then, find sinA, cosAand tanA.
2. The sides of a right angle triangle PQR are PQ = 7 cm, PR = 25 cm and ÐQ = 90o
respectively. Then find, tan P- tan R.
3. Inaright angle triangleABC withright angle at B, inwhich a =24 units, b =25 unitsand
Ð BAC = q. Then, find cos q and tan q.
4. If cosA=
12
13
, then find sinA and tanA (A<90o
).
5. If 3 tanA= 4, then find sinA and cosA.
6. In DABC and DXYZ, if Ð Aand Ð X are acute angles such that cosA= cos X then
show that Ð A= Ð X.
7. Given cot q =
7
8
, then evaluate (i)
(1 sin )(1 sin )
(1 cos )(1 cos )
∗ q , q
∗ q , q
(ii)
(1 sin )
cos
∗ q
q
8. In a right angle triangleABC, right angle is at B. If tanA= 3 , thenfind the value of
(i) sinA cos C + cos A sin C (ii) cos A cos C - sinA sin C
Q
P R
q
29
21

Class-X Mathematics
282
11.3 TRIGONOMETRIC RATIOS OF SOME SPECIFIC ANGLES
We alreadyknow about isosceles right angle triangle and right angle triangle with angles
30º, 60º and 90º.
Canwe find sin 30o
or tan 60o
or cos 45o
etc. withthe help ofthese triangles?
Does sin 0o
or cos 0o
exist?
11.3.1 TRIGONOMETRIC RATIOS OF 45O
In isosceles right angle triangleABC right angled at B
Ð A= Ð C = 45o
(why ?) and BC = AB (why ?)
Let’s assume the length ofBC =AB = a
Then, AC2
=AB2
+ BC2
(byPythagoras theorem)
= a2
+ a2
= 2a2
,
Therefore, AC = a 2
Using the definitionsoftrigonometricratios,
sin 45o
=
o
Length of theoppositesidetoangle45 BC 1
Length of hypotenuse AC 2 2
a
a
= = =
cos 45o
=
o
Length of theadjacentsidetoangle45 AB 1
Lengthof hypotenuse AC 2 2
a
a
= = =
tan 45o
=
o
o
Length of theoppositesidetoangle45 BC
1
AC
Lengthof theadjacentsidetoangle 45
a
a
= = =
Similarly, you can determinethe values of cosec 45o
, sec 45o
and cot 45o
.
AND 60O
Let us now calculate thetrigonometric ratios of 30o
and60o
.
Consider anequilateraltriangleABC. Sinceeachangle is
60o
in an equilateraltriangle, we have ÐA =ÐB =ÐC = 60o
and let the sides of equilateral triangle be
AB = BC = CA = 2a units.
Draw theperpendicular lineAD fromvertexAto
BC as shownin the adjacent figure.
C
A
B
A
B C
D
60º 60º
30º
2a 2a
a a
30º

Trigonometry 283
PerpendicularADis actsas“anglebisector ofangleA”and “bisector oftheside BC” in
the equilateraltriangleABC.
Therefore, Ð BAD = Ð CAD = 30o
.
Since point D divides the side BC intwo equalparts,
BD =
1 2a
BC
2 2
= = a units.
Consider right angletriangleABDintheabovegivenfigure.
We have AB = 2a and BD = a
Then AD2
=AB2
- BD2
by(Pythagoras theorem)
= (2a)2
- (a)2
= 3a2
.
Therefore, AD = a 3
Fromdefinitionsoftrigonometric ratios,
sin 60o
=
AD 3 3
AB 2 2
= =
a
a
cos 60o
=
BD 1
AB 2 2
= =
a
a
tan 60o
= 3 (how?)
Similarly, you can also determine the reciprocals, cosec 60o
, sec 60o
and cot 60o
.
DO THIS
Find the values of cosec 60o
, sec 60o
and cot 60o
.
TRY THIS
Find the values ofsin 30o
, cos30o
, tan 30o
, cosec 30o
, sec30o
and cot 30o
byusing
the ratio concepts.
AND 90O
Till now, we have discussed trigonometric ratios of 30o
, 45o
and 60o
. Now let us
determinethetrigonometric ratios ofangles 0o
and 90o
.

Class-X Mathematics
284
Suppose a segment AC of length r is
making anacute angle withrayAB. Height ofC
from B is BC. WhenAC leans more onAB so
that the angle made by it decreases, then what
happens to the lengths ofBC andAB ?
As the angleAdecreases, the height of
C fromAB raydecreasesand foot B is shiftedfromB to B1 and B2 and graduallywhenthe angle
becomes zero, height (i.e. opposite side ofthe angle) will also become zero (0) and adjacent
side would be equaltoAC i.e. lengthequalto r.
Let us lookat the trigonometric ratios
sin A=
BC
AC
and cos A =
AB
AC
IfA = 0o
then BC = 0 and AC = AB = r.
Thus, sin 0o
=
0
r
= 0 and cos 0o
=
r
r
= 1.
We know that tanA=
sin A
cosA
So, tan0o
=
o
o
sin0 0
0
1
cos0
= =
THINK AND DISCUSS
Discusswithyour friends about the following conditions:
1. What can you say about cosec 0o
=
1
sin0º
? Is it defined? Why?
C
A
B
r
q
C
A
B
C
A
B
C
Step (i) Step (ii)

Trigonometry 285
Step (i)
Step (ii) Step (iii)
2. What can you sayabout cot 0o
=
1
tan 0º
. Is it defined? Why?
3. sec 0o
= 1. Why?
Now let us see what happens when angle
made byAC with rayAB increases. When angleA
is increased, height ofpoint C increases and the foot
oftheperpendicularshiftsfromBto XandthentoY
and so on. In other words, we can say that the
height BC increases gradually, the angle on C gets
continuous increment and at one stage the angle
reaches 90o
. At that time, point B reachesAandAC
equalto BC.
So, when the angle becomes 90o
, base (i.e. adjacent side of the angle) would become
zero (0), theheight ofC fromABrayincreasesandit wouldbeequaltoACandthat isthelength
equalto r.
Now let ussee trigonometric ratios
sin A=
BC
AC
and cos A =
AB
AC
.
If A= 90o
thenAB = 0 and AC = BC = r.
Then , sin 90o
=
r
r
= 1 and cos 90o
=
0
r
= 0.
TRY THIS
Find the values for tan 90o
, cosec 90o
, sec 90o
and cot 90o
.

Class-X Mathematics
286
Now, let us see the values of trigonometric ratios ofallthe above discussed angles in
the formofa table.
Table 11.1
ÐA 0o
30o
45o
60o
90o
sinA 0
1
2
1
2
3
2
1
cosA 1
3
2
1
2
1
2
0
tanA 0
1
3
1 3 not defined
cotA not defined 3 1
1
3
0
sec A 1
2
3 2 2 not defined
cosecA not defined 2 2
2
3
1
THINK AND DISCUSS
Observe the above table
What can you say about the values of sinAand cosA, as the value of angleAincreases
from 0o
to 90o
?
IfA> B, then sinA> sin B. Is it true ?
IfA> B, then cosA> cos B. Is it true ? Discuss.
Example-4. In DABC, right angleis at B,AB = 5 cmand ÐACB = 30o
. Determine the lengths
of the sides BC andAC.
Solution : Given AB=5 cm and
ÐACB=30o
. To find the length ofside BC,
wewillchoosethetrignometricratioinvolving
BC and the given sideAB. Since BC is the
side adjacent to angle C and AB is the side
opposite to angle C.
Therefore,
AB
BC
= tan C
C
A
B
5
cm

Trigonometry 287
i.e.
5
BC
= tan 30o
=
1
3
which gives BC = 5 3 cm
Now, byusing the trigonometric ratios in DABC
5
sin 30
AC
=
o
1 5
2 AC
=
AC = 10 cm
Example-5. Achord of a circle of radius 6cm is making an angle 60o
at the centre. Find the
length of the chord.
Solution : Given that the radius ofthe circle is OA = OB = 6cm and
Ð AOB = 60o.
OC isheight from‘O’upon ABand it isanangle
bisector.
Then, Ð COB = 30o
.
Consider DCOB
sin 30o
=
BC
OB
1 BC
2 6
=
BC =
6
3
2
= .
But, length of the chord AB = 2BC
= 2 ´ 3 = 6 cm
Therefore, length ofthe chord = 6 cm.
O
C
A B
The first use of
the idea of ‘sine’
the way we use
it today was
given in the book
Aryabhatiyam
by Aryabhatta,
in 500 C.E.
Aryabhatta used
the word ardha-
jya for the half-
chord, which was shortened to jya or
jiva in due course. When the
Aryabhatiyam was translated into
Arabic, the word jiva was retained
as it is. The word jiva was translated
into sinus, which means curve, when
theArabic version was translated into
Latin. Soon the word sinus, also used
as sine, became common in
mathematical texts throughout
Europe. An English Professor of
astronomy Edmund Gunter (1581–
1626), first used the abbreviated
notation ‘sin’.

Class-X Mathematics
288
Example-6. In DPQR, right angle is at Q, PQ = 3 cmand PR = 6 cm. Determine ÐQPR and
ÐPRQ.
Solution : Given PQ = 3 cm and PR = 6 cm
Therefore,
PQ
PR
= sin R
or sin R =
3 1
6 2
=
So, ÐPRQ = 30o
and therefore, ÐQPR = 60o
(why?)
THINK AND DISCUSS
If one of the sides and any other part (either an acute angle or anyside) ofa right angle
triangle is known, the remaining sides and angles ofthe triangle canbe determined. Do
you agree?Explainwith an example.
Example-7. Ifsin(A-B)=
1
2
,cos(A+B)=
1
2
, where0o
<A+B<90o
andA>B,findAandB.
Solution : Since sin (A- B) =
1
2
, therefore,A- B = 30o
(why?)
Also, since cos (A+ B) =
1
2
, therefore,A+ B = 60o
(why?)
Solving the above equations, we get :A= 45o
and B = 15o
. (How?)
EXERCISE - 11.2
1. Evaluate the following.
(i) sin 45o
+ cos 45o
(ii)
o
o o
cos45
sec30 cosec60
+
(iii)
o o o
o o o
sin 30 tan45 cosec60
cot 45 cos60 sec30
+ -
+ -
(iv) 2 tan2
45o
+ cos2
30o
- sin2
60o
(v)
2 o 2 o
2 o 2 o
sec 60 tan 60
sin 30 cos 30
-
+
2. Choose the right option and justify your choice-
(i)
o
2 o
2tan30
1 tan 45
+
=
(a) sin 60o
(b) cos 60o
(c) tan 30o
(d) sin 30o
P
R
Q

Trigonometry 289
(ii)
2 o
2 o
1 tan 45
1 tan 45
-
+
=
(a) tan 90o
(b) 1 (c) sin 45o
(d) 0
(iii)
o
2 o
2tan30
1 tan 30
-
=
(a) cos 60o
(b) sin 60o
(c) tan 60o
(d) sin 30o
3. Evaluate sin 60o
cos 30o
+ sin 30o
cos 60o
. What is the value of sin(60o
+ 30o
). What
can you conclude ?
4. Is it right to say that cos(60o
+ 30o
) = cos 60o
cos30o
- sin 60o
sin 30o
.
5. Inright angletriangleDPQR, right angleis atQ,PQ=6cmand ÐRPQ= 60o
. Determine
the lengths ofQR and PR.
6. In DXYZ, right angle is at Y, YZ = x, and XZ = 2x. Then determine Ð YXZ and
ÐYZX.
7. Is it right to say that sin (A + B) = sin A + sin B? Justify your answer.
THINK AND DISCUSS
For which value of an acute angle q, (i)
cos cos
4
1 sin 1 sin
q q
+ =
- q + q
is true?
For which value of 0o
< q < 90o
, above equation is not defined?
11.4 TRIGONOMETRIC RATIOS OF COMPLEMENTARY ANGLES
We alreadyknow that two angles are said to be complementary, iftheir sumis equalto
90o
. Consider a right angle triangleABC with right angle at B. Are there anycomplementary
angles inthistriangle?
Since angle B is 90o
, sumofother two angles must be
90o
. (Q Sum of angles in a triangle 180o
)
Therefore, Ð A + Ð C = 90o
. Hence Ð A and Ð C
are complementaryangles.
Let us assume that Ð A = x, then for angle x, BC is
opposite side and AB is adjacent side.
C
A B

Class-X Mathematics
290
sin x =
BC
AC
cos x =
AB
AC
tan x =
BC
AB
cosec x =
AC
BC
sec x =
AC
AB
cot x =
AB
BC
If Ð A + Ð C = 90o
, then we have Ð C = 90o
- Ð A
Since Ð A = x, we have Ð C = 90o
- x
Let us look at what would be “Opposite side” and “Adjacent side” of the angle
(90o
- x)in the triangleABC.
sin(90o
- x) =
AB
AC
cos(90o
- x) =
BC
AC
tan(90o
- x) =
AB
BC
Cosec(90o
- x) =
AC
AB
sec(90o
- x) =
AC
BC
cot(90o
- x) =
BC
AB
Now, if we compare the ratios of angles x and (90o
- x) from the above values of
different triginometricratios, wegetthefollowingrelations:
sin(90o
- x) =
AB
AC
= cos x and cos(90o
- x) =
BC
AC
= sin x
tan(90o
- x) =
AB
BC
= cot x and cot(90o
- x) =
BC
AB
= tan x
cosec(90o
- x) =
AC
AB
= sec x and sec(90o
- x) =
AC
BC
= cosec x
THINK AND DISCUSS
Check and discuss the above relations in the case of angles between 0º and 90º,
whether theyhold for these angles or not?
So, sin (90o
- A) = cosA cos (90o
- A) = sinA
tan (90o
-A) = cot A cot (90o
-A) = tanA
sec (90o
- A) = cosec A and cosec (90o
- A) = sec A
Now, let us consider some examples.

Trigonometry 291
Example-8. Evaluate
o
o
sec35
cosec55
Solution : cosec A = sec (90o
- A)
cosec 55o
= cosec (90o
- 35o
)
cosec 55o
= sec 35o
Now
o o
o o
sec35 sec35
cosec55 sec35
= = 1
Example-9. If cos7A= sin(A- 6o
), where 7Ais an acute angle, find the value ofA.
Solution : Given cos 7A = sin(A - 6o
) ...(1)
sin (90 - 7A) = sin (A - 6o
)
since (90 - 7A) & (A - 6o
) are both acute angles,
therefore
90o
- 7A = A - 6o
8A = 96o
which gives A= 12o
.
Example-10. If sinA= cos B, then prove that A+ B = 90º.
Solution : Given that sinA= cos B ...(1)
We know cos B = sin (90o
- B).
We can write (1) as
sinA= sin (90o
- B)
Since A, B are acute angles, A= 90o
- B
Þ A + B = 90o
.
Example-11. Express sin 81o
+ tan 81o
in terms of trigonometric ratios of angles between
0o
and 45o
.
Solution : We can write sin 81o
= sin(90o
- 9o
) = cos 9o
and tan 81o
= tan(90o
- 9o
) = cot 9o
Then, sin 81o
+ tan 81o
= cos 9o
+ cot 9o

Class-X Mathematics
292
Example-12. IfA,BandCareinterioranglesoftriangleABC,thenshowthat
B C A
sin ( ) cos
2 2
+
=
Solution : Given A, B and C are angles oftriangle ABC then
A + B + C = 180o
.
On dividing the above equation by 2on both sides, we get
o
A B C
90
2 2
+
+ =
o
B C A
90
2 2
+
= -
On taking sinratio onbothsides
o
B C A
sin sin 90
2 2
+
æ ö æ ö
= -
ç ÷ ç ÷
è ø è ø
B C A
sin cos
2 2
+
æ ö
=
ç ÷
è ø . Hence proved.
EXERCISE 11.3
1. Evaluate
(i)
o
o
tan36
cot54
(ii) cos12o
- sin78o
(iii) cosec 31o
- sec 59o
(iv) sin 15o
sec 75o
(vi) tan 26o
tan64o
2. Showthat
(i) tan 48o
tan 16o
tan 42o
tan 74o
= 1
(ii) cos36o
cos 54o
- sin360
sin 54o
= 0.
3. Iftan 2A= cot(A- 18o
), where 2Ais an acute angle. Find the value ofA.
4. If tanA= cot B whereAand B are acute angles, prove that A+ B = 90o
.
5. IfA, B and C are interior angles ofa triangleABC, then show that
A B C
tan cot
2 2
+
æ ö
=
ç ÷
è ø
6. Expresssin75o
+cos65o
intermsoftrigonometricratiosofanglesbetween 0o
and45o
.

Trigonometry 293
11.5 TRIGONOMETRIC IDENTITIES
We know that an identityisthat mathematicalequation whichis true for allthe values of
the variablesinthe equation.
For example (a + b)2
= a2
+ b2
+ 2ab is an identity.
In the same way, an identity equation having trigonometric ratios of an angle is called
trigonometric identity. It is true for allthe values ofthe anglesinvolved init.
Here, we willderive a trigonometric identityand remaining would be based on that.
Consider aright angletriangleABCwithright angle at B.
FromPythagoras theorem
We have AB2
+ BC2
= AC2
....(1)
Dividing each termbyAC2
, we get
2 2 2
2 2 2
AB BC AC
AC AC AC
Þ + =
i.e.,
2 2 2
AB BC AC
AC AC AC
é ù é ù é ù
+ =
ê ú ê ú ê ú
ë û ë û ë û
i.e., (cosA)2
+ (sinA)2
= 1
Here, we generallywrite cos2
Ainthe place of(cosA)2
i.e., (cosA)2
is written as cos2
A(Do not write cosA2
)
above equation is cos2
A+ sin2
A= 1
We have given an equation having avariable parameterA(angle) and above equation is
true for all the value ofA. Hence the above equation is a trigonometric identity.
Therefore, we have trigonometric idenity
cos2
A + sin2
A = 1.
Let uslookat another trigonometric idenity
From equation(1) we have
AB2
+ BC2
= AC2
2 2 2
2 2 2
AB BC AC
AB AB AB
Þ + = (Dividing each term by AB2
)
C
A
B
C
A
B

Class-X Mathematics
294
2 2 2
AB BC AC
AB AB AB
æ ö æ ö æ ö
+ =
ç ÷ ç ÷ ç ÷
è ø è ø è ø
i.e., 1 + tan2
A = sec2
A
Similarly, on dividing (1) by BC2
, we get cot2
A+ 1 = cosec2
A.
Byusingaboveidentities,wecanexpresseachtrigonometricratio intermsofanotherratio.
If weknow the value ofaratio, we can findallother ratios byusing these identities.
THINK AND DISCUSS
Aretheseidentitiestrueonlyfor00
<A<900
? Ifnot,forwhichothervaluesofAtheyaretrue?
l sec2
A - tan2
A= 1 l cosec2
A - cot2
A = 1
DO THIS
(i) If sin C =
15
,
17
then find cos C. (ii) If tan x =
5
,
12
then find sec x.
(iii) If cosec q =
25
7
, then find cot q.
TRY THIS
Evaluate the following andjustifyyouranswer.
(i)
2 o 2 o
2 o 2 o
sin 15 sin 75
cos 36 cos 54
+
+
(ii) sin 5o
cos 85o
+ cos5o
sin 85o
(iii) sec 16o
cosec 74o
- cot 74o
tan 16o
.
Example-13. Show that cot q + tan q = sec q cosec q.
Solution : LHS = cot q +tan q
cos sin
sin cos
q q
= +
q q
(why ?)
2 2
cos sin
sin cos
q + q
=
q q

Trigonometry 295
1
sin cos
=
q q (why ?)
1 1
cosec sec
sin cos
= = q q
q q
Example-14. Show that tan2
q + tan4
q = sec4
q - sec2
q
Solution : L.H.S. = tan2
q + tan4
q
= tan2
q (1+ tan2
q)
= tan2
q . sec2
q (Why ?)
= (sec2
q - 1) sec2
q (Why ?)
= sec4
q - sec2
q = R.H.S
Example-15. Prove that
1 cos
1 cos
+ q
- q
= cosec q + cot q; 0 < q < 900
.
Solution : LHS =
1 cos
1 cos
+ q
- q
(multiplynumerator anddenominator by (1 cos )
+ q )
1 cos 1 cos
.
1 cos 1 cos
+ q + q
=
- q + q
2
2
(1 cos )
1 cos
+ q
=
- q
2
2
(1 cos )
sin
+ q
=
q
(Why ?)
1 cos
sin
+ q
=
q
1 cos
cosec cot R.H.S.
sin sin
q
= + = q + q =
q q
EXERCISE 11.4
1. Evaluate the following :
(i) (1 + tan q + sec q) (1 + cotq - cosec q)
(ii) (sin q + cos q)2
+ (sin q - cos q)2
(iii) (sec2
q - 1) (cosec2
q -1)

Class-X Mathematics
296
2. Show that (cosec q - cot q)2
=
1 cos
1 cos
- q
+ q
.
3. Show that
1 sin A
secA tan A
1 sin A
+
= +
-
.
4. Show that
2
2
2
1 tan A
tan A
cot A 1
-
=
-
.
5. Show that
1
cos tan .sin
cos
- q = q q
q
.
6. SimplifysecA (1 - sinA) (secA+ tanA).
7. Prove that (sinA+ cosecA)2
+ (cosA+ secA)2
= 7 + tan2
A+ cot2
A.
8. Simplify (1 - cos q) (1 +cosq) (1 + cot2
q).
9. If secq + tan q = p, then what is the value of secq - tan q ?
10. If cosec q + cot q = k, then prove that cos q =
2
2
k 1
k 1
-
+
.
OPTIONAL EXERCISE
1. Prove that cot cos cosec 1
cot cos cosec 1
q - q q -
=
q + q q +
.
2. Prove that
sin cos 1 1
sin cos 1 sec tan
q - q +
=
q + q - q - q
[use the identity sec2
q = 1 + tan2
q].
3. Prove that (cosecA- sinA) (secA- cosA) =
1
tan A cot A
+
.
4. Prove that
2
1 secA sin A
secA 1 cosA
+
=
-
.
5. Show that
2
2
2
2
1 tan A 1 tan A
tan A
1 cotA
1 cot A
æ ö
+ +
æ ö
= =
ç ÷
ç ÷ è ø
-
+
è ø
.
6. Prove that
(secA 1) (1 cosA)
(secA 1) (1 cos A)
, ,
<
∗ ∗
.

Trigonometry 297
1. In a right angle triangleABC, withright angle at B,
Side oppositeto angle A
sin A
Hypotenuse
= ,
Side adjacent to angle A
cosA
Hypotenuse
=
2.
1 1 sin A 1
cosecA ; secA ; tanA ; tan A
sin A cosA cosA cot A
= = = =
3. Ifone ofthe trigonometric ratios ofanacute angle is known, the remaining trignometric
ratios oftheangle can be determined.
4. The values ofthetrigonometric ratios for angle 0o
, 30o
, 45o
, 60o
and 90o
.
5. The value of sinAor cosAnever exceeds 1, whereas the value of secAor cosecAis
always greater thanor equalto 1.
6. sin (90o
- A) = cos A, cos (90o
- A) = sinA
tan (90o
- A) = cot A, cot (90o
- A) = tanA
secA (90o
- A) = cosec A, cosec (90o
- A) = sec A
7. sin2
A + cos2
A= 1
sec2
A - tan2
A= 1 for 0o
< A< 90o
cosec2
A - cot2
A = 1 for (0o
< A< 90o
)

298 Class-X Mathematics
Applications of
Trigonometry
12
12.1 INTRODUCTION
You have studied in socialstudiesthat the highest mountainpeak in the world is Mount
Everest and its height is 8848 meters.
KuntalawaterfallinAdilabad district isthehighest naturalwaterfallinTelangana. Itsheight
is 147 feet.
How were theseheights measured? Canyou measurethe height ofyour schoolbuilding or
the tallest tree inor around your school?
Let us understand through some examples. Vijaya
wants to findthe height ofapalmtree. She triesto locate the
topmost point ofthe tree.She also imaginesaline joiningthe
top most point and her eye.
This line is called “line of sight”. She also imagines a
horizontalline, fromher eye to the tree.
Here,“thelineofsight”,“horizontalline”and“thetree”
formaright angletriangle.
To find the height of the tree, she needs to find a side
and ananglein thistriangle.
“The line ofsight isabovethe horizontalline andangle
betweenthelineofsight andthe horizontallineiscalled angle
of elevation”.
q
L
i
n
e
o
f
s
i
g
h
t
Angle of
elevation

299
Applications of Trigonometry
Supposeyouarestanding
on the top of your school
buildingandyouwant tofindthe
distance of borewell from the
building on which you are
standing. For that, you have to
observe the base of the
borewell.
Then,thelineofsightfrom
your eyeto thebaseofborewell
is belowthe horizontalline fromyour eye.
Here, “theanglebetweentheline ofsight andhorizantalline is calledangleofdepression.”
Trigonometry hasbeen used bysurveyors for centuries. TheyuseTheodolites to measure
anglesofelevationor depressioninthe processofsurvey. Innineteenthcentury,two large
TheodoliteswerebuiltbyBritishIndiaforthesurveyingproject“greattrigonometricsurvey”.
During the surveyin 1852, the highest mountainpeak inthe world was discovered in the
Himalayas. From the distance of 160 km, the peak was observed from six different
stationsand the height ofthe peak wascalculated. In1856, this peak wasnamed after Sir
George Everest, who had commissioned and first used the giant Theodolites. Those
theodolites are kept inthe museumofthe SurveyofIndia inDehradun for display.
12.2 DRAWING FIGURES TO SOLVE PROBLEMS
When we want to solve the problems of heights and distances, we should consider the
following:
(i) Alltheobjectssuchas towers, trees,buildings, ships, mountainsetc. shallbeconsidered
as linear for mathematicalconvenience.
(ii) Theangleofelevationorangleofdepressionisconsideredwithreferenceto thehorizontal
line.
(iii) The height ofthe observer is neglected, ifit isnot givenin the problem.
Whenwe tryto find heightsand distancesat anangle ofelevation ordepression, we need
to visualise geometrically. To find heights and distances, we need to draw figures and with the
help of these figures we can solve the problems. Let us see some examples.
Example-1. The topofa clock toweris observed at angle ofelevationof o
a and thefoot ofthe
tower is at the distance of d metersfromthe observer. Draw the diagramfor this data.
q
Horizontal line
Angle of depression
Line of sight

q1
q2
q2
q2
q1
q2
E
D
A
B
C
h
b
b
Solution : The diagrams are as shown below :
Example-2. Rinky observes a flower on the ground from the balcony of the first floor of a
building at an angle of depression bo
. The height of the first floor of the building is x meters.
Draw the diagramfor this data.
Solution :
Here ÐDAC = ÐACB = b (why?)
Example-3. Alarge balloon has been tiedwith a rope and it is floating inthe air. Aperson has
observed the balloonfromthe topofa building at angle ofelevation ofq1 and foot of theropeat
anangle ofdepressionof q2. The height ofthebuilding is h feet. Drawthediagramforthisdata.
Solution : We cansee that
ÐBDA= ÐDAE. (Why?)
d
a
d
a
A
B C
b
b
B
x
C
A
D

301
DO THIS
1. Draw diagramforthefollowing situations :
(i) A person is flying a kite at an angle of elevation a and the length of thread
fromhis hand to kite is ‘l’.
(ii) A person observes two banks of a river at angles of depression q1 and q2
(q1 < q2 ) fromthe top ofa treeofheight h which isat a side ofthe river. The
width oftheriver is ‘d’.
THINK AND DISCUSS
1. You are observing top ofyour schoolbuilding at an angle ofelevation afroma point
which is at d meter distance fromfoot ofthe building.
Whichtrigonometricratio would you liketoconsider to findtheheightofthebuilding?
2. A ladder of length x meter is leaning against a wall making angle q withthe ground.
Whichtrigonometricratio wouldyou liketo consider to findtheheight ofthepoint on
the wallat whichthe ladder istouching?
Tillnow, we havediscussedhowto draw diagramsasper the situationsgiven. Now, we
shalldiscuss how to find heightsand distances.
Example-4. Aboyobserved the top ofan electric pole at an angle ofelevation of 60º when the
observation point is 8 meters awayfromthe foot ofthe pole. Find the height ofthe pole.
Solution : Fromthe figure, intriangle OAB
OB = 8 meters and
ÐAOB = 60º.
Let height ofthepole =AB = h meters
(we know the adjacent side and we need to find the opposite side of ÐAOB in the triangle
DOAB. Hence we need to consider the trigonometric ratio “tan” to solve the problem).
tan 60º =
AB
OB
3
8
h
= 8 3 .
h m
=
A
B
O
60°
h
8 m

P O
B
A
45º
500 m
45º
Example-5. Rajender observes aperson standing on theground froma helicopter at an angle of
depression 45º. If the helicopter flies at a height of 500 meters from the ground, what is the
distance ofthe person fromRajender?
Solution : Fromthe figure, intriangle OAB
OA = 500 meters and
POB
Ð = OAB
Ð = 45º (why ?)
OB = distance of the person fromRajender = x.
(we know the oppositeside of OBA
Ð and we need to find hypotenuse OB in thetriangle OAB.
Hence, we need to consider the ratio “sin”.)
sin 45º =
OA
OB
1 500
2 x
=
500 2
x = meters
(The distance fromthe person to Rajendar is 500 2 m)
EXERCISE - 12.1
1. A towerstands verticallyon theground. Froma point which is 15 meterawayfromthe
foot ofthe tower, the angleofelevationofthetop ofthe toweris 45º. What is the height
ofthe tower?
2. A tree breaks due to stormand the broken part bends so that the top ofthe tree touches
the ground bymaking 30º angle withthe ground. The distance between the foot ofthe
tree and the top of thetree on the groundis 6m. Find the height ofthe tree before falling
down.
3. A contractor wants to set up a slide for the children to playinthe park. He wants to set
it up at the height of 2 mand bymaking anangle of30º with the ground. What should be
the lengthofthe slide?
4. Length of the shadow of a 15 meter high pole is 15 3 meters at 8 O’clock in the
morning. Then,what istheangleofelevationofthe Sunrays withthe groundat thetime?
5. You want to erect a pole ofheight 10 mwith the support ofthree ropes. Eachrope has
to make an angle 30º with the pole. What should be the lengthofthe rope?
x

303
6. Suppose you are shooting an arrow from the top ofa building at an height of6 mto a
target onthe ground at anangle ofdepression of 60º. What is the distance between you
and the object?
7. An electrician wants to repair an electric connection on a pole of height 9 m. He needs
to reach 1.8 mbelow the top ofthe pole to do repair work. What should be the length
of theladder which he shoulduse, when he climbsit at anangle of 60º withthe ground?
What willbe the distance between foot ofthe ladder and foot ofthe pole?
8. Aboat hasto cross a river. It crosses the river bymaking anangle of60º withthe bank
of the river due to the stream of the river and travels a distance of600m to reach the
another sideofthe river. What is the width ofthe river?
9. An observer ofheight 1.8 mis13.2 mawayfroma palmtree. The angle ofelevation of
the top ofthe tree fromhis eyes is 45º. What is the height ofthe palmtree?
10. Intheadjacent figure,
AC = 6 cm, AB = 5
cm and BAC
Ð =
30º. Find the area of
thetriangle.
12.3 SOLUTION FOR TWO TRIANGLES
We have discussed the
solutionofaone triangle problem.
What will be the solution if there
are two triangles?
Suppose you are standing
on one side of a tree.You want to
find the height of a tree observing
the tree from different points of
observations.
How can you do this?
Supposeyouareobservingthetop
of the palm tree at an angle of
elevation 45º. The angle of
elevationchangesto30ºwhenyou
move 11 mawayfromthe tree.
45º
30º
A C
B
D
E1
E
A
30º
6 cm
5 cm
C
B

D
B
A
30º
h
45º
C
11 m
E
F
Let us see how we can find height ofthe tree.
Fromfigure, we have
AB = 11 m
DAC
Ð = 30º
DBC
Ð = 45º
Let the height of the palmtree CD = hmeters
and length of BC = x.
Then AC = 11 + x.
Fromtriangle BDC,
tan 45º =
DC
BC
1
h
x h
x
< Þ < ...(1)
FromtriangleADC,
tan 30º =
DC
AC
1
11
3
h
x
=
+
11
3
x
h
+
=
11
3 3
h
h = +
11
3 3
h
h- =
( )
3 1 11
3 3
h
-
=
( )
11
3 1
h =
-
meters.
Note : Totalheight ofthe palmtree is CD + CE whereCE=AF, whichisthe height ofthe girl.

305
Example-6. Two menoneither side ofatemple of30 meter height observe its topat the angles
of elevation30º and 60º respectively. Find the distance between the two men.
Solution : Height ofthe temple BD =30 meter.
Angle of elevation ofone person ÐBAD= 30º
Angle of elevation of another person ÐBCD= 60º
Let the distance between the first person and the temple, AD = x and distance between the
second person and the temple, CD = d
FromDBAD From DBCD
tan 30º =
BD
AB
tan 60º =
BD
d
1 30
3 x
< 30
3
d
<
30 3
x < .......... (1)
30
3
d < .......... (2)
from (1) and (2) distance between the persons = BC + BA = x + d
30 30 4 120
30 3 40 3
3 3 3
≥
< ∗ < < < meter
Example-7. Astraight highwayleadsto the foot ofa tower. Ramaiahstanding at the top ofthe
tower observesa car at anangle ofdepression 30º. The caris approaching the foot ofthe tower
with a uniformspeed. Sixseconds later, the angle ofdepression of the car is found to be 60º.
Find the time taken bythe car to reach the foot of the tower fromthis point.
Solution :
Let the distance travelled by the car in 6 seconds =AB = x meters
Heights ofthe tower CD = h meters
The remaining distance to be travelledbythe car BC = d meters
andAC = AB + BC = (x + d) meters
ÐPDA = ÐDAB = 30º (why?)
ÐPDB = ÐDBC = 60º (why?)
From DBCD
x
30º 60º
d
A
B
C
D

P
D
B
A
30º
h
60º
C
d
30º
60º
CD
tan 60º
BC
=
3
h
d
=
3
h d
= ...(1)
From DACD
CD
tan30º
AC
=
1
( )
3
h
x d
=
+
( )
3
x d
h
+
= ...(2)
From (1) & (2), we have
3
3
x d
d
+
=
x + d = 3d
x = 2d
2
x
d =
Time takento travel‘x’meters = 6 seconds.
Time takento travelthe distance of‘d’meters
i.e.,
2
x
meters is 3 seconds.
EXERCISE - 12.2
1. A TV tower stands vertically on the side of a road. From a point on the other side
directlyopposite to the tower, the angle ofelevation ofthe top oftower is 600
. From
another point 10 mawayfromthispoint, onthe line joining this point to the foot ofthe
tower, the angle ofelevationofthe top ofthetower is 30º. Find the height ofthe tower
and the widthofthe road.

307
2. A1.5 mtallboyis looking at the top ofatemple whichis 30 meter in height froma point
at certain distance. The angle ofelevation fromhis eye to the top ofthe crown ofthe
temple increases from 30º to 60º as he walks towards the temple. Find the distance he
walked towardsthe temple.
3. A statue stands on the top ofa 2m tallpedestal. Froma point on the ground, the angle
ofelevation ofthe topofthe statue is 60ºand fromthe same point, theangle ofelevation
ofthe top of the pedestal is 45º. Find the height ofthe statue.
4. Fromthe top ofa building, the angle ofelevationofthe top ofa celltower is 60º and the
angle ofdepressionto its foot is 45º. Ifdistance ofthe building fromthetower is 7m,
thenfind theheight ofthe tower.
5. A wire oflength 18 mhad been tied with electric pole at an angle ofelevation 30º with
the ground. Because it was covering a long distance, it was cut and tied at an angle of
elevation 60º withthe ground. How much length ofthe wire was cut?
6. The angle ofelevation ofthe top ofa building from the foot ofthe tower is 30º and the
angle ofelevationofthetopofthe tower fromthe foot ofthebuilding is60º. Ifthe tower
is 30 mhigh, find the height of thebuilding.
7. Two polesofequalheightsarestanding opposite to eachother oneither sideofthe road,
whichis120 feet wide. Froma point betweenthemon the road, theangles ofelevation
of thetop of the polesare 60º and 30º respectively. Find the height ofthe poles and the
distancesofthe point fromthe poles.
8. The angles of elevation of the top of a tower fromtwo points at a distance of4 mand
9 m,find the height ofthe tower fromthe base ofthe towerand in the samestraight line
withitarecomplementary.
9. The angleofelevationofajet plane fromapointAon the ground is 60º.After a flight of
15 seconds, the angle ofelevation changes to 30º. Ifthe jet plane is flying at a constant
height of1500 3 meter, find the speed of thejet plane. ∋ (
3 1.732
<
10. The angle ofelevation ofthe top ofa tower from the foot ofthe building is 30º and the
angle ofelevationofthe topofthe building fromthe foot ofthetower is 60º. What is the
ratio ofheightsoftower and building.
OPTIONAL EXERCISE
1. A 1.2 mtallgirlspots a balloon moving withthe wind in a horizontalline at a height of
88.2 mfromthe ground. The angleofelevationofthe balloonfromthe eyes ofthe girlat
aninstantis60º.After sometime, theangleofelevationreducesto30º.Findthedistance
travelled bythe balloonduring the interval.

2. The anglesof elevationofthetopofalighthousefrom3boatsA,B andC inastraight line
ofsame sideofthe light houseare a,2a,3arespectively. Ifthedistancebetweentheboats
Aand Band the boatsB andC are x and y respectivelyfind the height ofthe light house?
3. Inner part of acupboard is in the cuboidicalshapewithits length, breadthand height in
the ratio 1 : 2 : 1. What is the angle made bythe longest stick whichcan be inserted
cupboard withits base inside.
4. An iron spericalballofvolume 232848 cm3
has been melted and converted into a cone
with the verticalangle of 120o
. What are its height and base?
5. Show that the area of an Issosceles triangle is A = a2
Sin q Cos q
where a isthe length ofoneofthe two equalsides and qisthe measure
of one oftwo equal angles
6. Aright circular cylindricaltower with height ‘h’and radius ‘r’, stands onthe ground.
Let ‘p’be a point in the horizontalplane ground andABC be the semi-circular edge of
the top ofthe tower such that Bis the point in it nearest to p. The angles ofelevation of
the pointsAand B are 45o
and 60o
respectively. Show that
3(1 3)
2
h
r
+
= .
Inthis chapter, we have studied thefollowing points:
1. (i) Thelineofsightisthelinedrawnfromtheeyeofanobserverto apoint ontheobject
being viewed by the observer.
(ii) The angle ofelevation ofthe object viewed, is theangle formed bythe line ofsight
with the horizontalwhen it isabove the horizontal level, i.e., the case whenwe raise
our head to look at the object.
(iii) The angleofdepression ofanobject viewed, istheangle formed bythe line ofsight
withthehorizontalwhenit isbelowthehorizontallevel,i.e.,thecasewhenwelower
our head to look at the object.
2. The height or length of an object or the distance between two distant objects can be
determined with thehelp of trigonometric ratios.
Suggested Projects
Find the heights and distances
l Usingclinometer -find the height ofatower/ tree/ building.
q q
a a

13.1 INTRODUCTION
Kumar and Sudha were walking together to playa carromsmatch:
Kumar: Do you think that we would win?
Sudha : There are 50 percent chances for that.We maywin.
Kumar: How do you say 50 percent?
Do youthink Sudha isright inher statement?
Is her chanceofwining 50%?
In this chapter, we study about suchquestions. We also discuss words like ‘probably’,
‘likely’, ‘possibly’, etc. and how to quantifythese. In classIX we studied about events that are
extremelylikelyand in fact, are almost certain and thosethat are extremelyunlikelyand hence
almost impossible. We also talked about chance, luck and the fact that an event occurs once
does not meanthat it would happeneachtime. In thischapter, we tryto learnhow the likelihood
of anevent can be quantified.
This quantificationinto a numerical measure is referred to as finding 'Probability'.
13.1.1 WHAT IS PROBABILITY
Consider anexperiment: Anormalcoinwastossed1000 times.Head turnedup455times
and tailturnedup 545 times. Ifwe tryto find the likelihood ofgettingheads we maysayit is 455
out of 1000 or
455
1000
or 0.455.
This estimation of probability is based on the results of an actual
experiment oftossingacoin1000times.Theseestimatesarecalledexperimental
or empiricalprobabilities. Infact, allexperimentalprobabilities are based on
the resultsofactualexperiments andanadequate recording ofwhat happens
in eachof the events. These probabilitiesare only'estimations'.If we perform
the same experiment for another 1000 times, we may get slightly different
data, givingdifferent probabilityestimate.
Probability
13

Class-X Mathematics
310
Many persons fromdifferent parts of the world have done this kind ofexperiment and
recorded the number of heads that turned up.
For example, the eighteenth century French naturalist Comte de Buffon, tossed a coin
4040 times and got 2048 times heads. The experimental probabilityofgetting a head, in this
case, was
2048
4040
i.e., 0.507.
J.E.Kerrich, fromBritain,recorded5067headsin10000tossesofacoin. The experimental
probabilityofgetting a head, in this case, was
5067
10000
= 0.5067. Statistician KarlPearson spent
some more time, making 24000 tosses of a coin. He got 12012 times heads, and thus, the
experimentalprobability ofa head obtained byhimwas 0.5005.
Now, suppose we ask, 'What willbe the experimentalprobabilityofgetting a head, ifthe
experiment iscarried onup to,say, one milliontimes?Or 10 milliontimes?You would intuitively
feelthat as the number of tosses increases, the experimentalprobabilityof a head (or a tail) may
settledowncloserandcloser to the number 0.5, i.e.,
1
2
.Thismatchesthe‘theoreticalprobability’
ofgetting a head (or getting a tail), about whichwewilllearnnow.
This chapter is an introduction to the theoretical (also called classical) probability of an
event. Now we discuss simple problems based on this concept.
13.2 PROBABILITY - A THEORETICAL APPROACH
Let us considerthe following situation: Suppose a ‘fair’coin is tossed at random.
Whenwespeak ofa coin,we assume it to be 'fair', that is, it issymmetricalso that there is
no reasonfor it to comedown moreoftenonone side thanthe other. We callthispropertyofthe
coin as being 'unbiased'. By the phrase 'randomtoss', we mean that the coin is allowed to fall
freely without any bias or interference. These types of experiments are random experiments
(Here wedismiss the possibilityofits 'landing' on its edge, which maybepossible, for example,
ifit falls onsand). We refer to this bysaying that the outcomes, head and tail, areequallylikely.
Forbasicunderstandingofprobability,inthischapter,wewillassumethatalltheexperiments
have equallylikelyoutcomes.
Now, we know that the experimentalor empiricalprobabilityP(E) ofanevent E is
P(E) =
Number of trials in which the event happened
Total number of trials

Probability 311
DO THIS
a. Outcomes ofwhichofthe following experimentsare equallylikely?
1. Getting a digit 1, 2, 3, 4, 5 or 6 when a die is rolled.
2. Selecting a different colour ball from a bag of 5 red balls, 4 blue balls and 1 black
ball.
3. Winning in agame ofcarrom.
4. Units place of a two digit number selected may be 0, 1, 2, 3, 4, 5, 6, 7, 8 or 9.
5. Selecting a different colour ball from a bag of 10 red balls, 10 blue balls and 10
black balls.
6. Raining ona particular dayofJuly.
b. Are the outcomes ofeveryexperiment equallylikely?
c. Giveexamplesof5experimentsthat haveequallylikelyoutcomesandfivemoreexamples
that do not have equallylikelyoutcomes.
ACTIVITY
(i) Take any coin, toss it, 50 times, 100 times, 150 times and count the number of times a
head and a tailcome up seperately.Record your observations inthe following table:-
S. Number of Number of Probability of Number of Probability of
No. experiments heads head tails tails
1. 50
2. 100
3. 150
What do you observe? Obviously, as the number of experiments increases, probability
of head or tailreaches50% or
1
2
. Thisempiricalinterpretationofprobabilitycanbe applied
to everyevent associated with anexperiment that canbe repeated a large number of times.
Probability and Modelling
The requirement of repeating an experiment has some limitations, as it may be very
expensive or unfeasible in many situations. Ofcourse, it worked wellin the experiments of
tossing the coin or throwing adice. But how about repeating the experiment oflaunching a
satellite inorder to compute the empiricalprobabilityofits failure during launching, or the

Class-X Mathematics
312
repetition of the phenomenon of an earthquake to compute the empiricalprobabilityof a
multi-storeyedbuildinggettingdestroyedinanearthquake?Forfindingtheseprobabilitieswe
calculate models ofbehaviour and use themto estimate behaviour and likelyoutcomes. Such
models are complex and are validated by predictions and outcomes. Forecast of weather,
result ofan election, population demography, earthquakes, crop production etc. are allbased
on such modelsand their predictions.
“Theassumptionofequallylikelyoutcomes”(whichisvalid inmanyexperiments,asintwo
of the examples seen, of a coin and of a dice) is one of the assumption that leads us to the
following definitionofprobabilityofanevent.
Thetheoreticalprobability(also calledclassicalprobability)ofaneventT,writtenasP(T),
is defined as
P(T) =
Number of outcomes favourable to T
Number of all possible outcomes of the experiment
where weassumethat theoutcomesoftheexperiment are equallylikely.Weusuallysimply
refer to theoreticalprobabilityasProbability.
The definition of probability was given byPierre Simon Laplace
in 1795.
Probabilitytheoryhadits originin the16th
centurywhen anItalian
physicianandmathematicianJ.Cardanwrotethefirstbookonthesubject,
The Book on Games of Chance. James Bernoulli (1654 -1705),A. De
Moivre (1667-1754), and Pierre Simon Laplace are among those who
made significant contributionsto this field. Inrecent years, probability
has been used extensively in many areas such as biology, economics,
genetics, physics, sociologyetc.
13.3 MUTUALLY EXCLUSIVE EVENTS
Ifa coinis tossed, we get a head or atail, but not both. Similarly, ifwe select a student of
a highschoolthat he/ shemaybelong to one of either 6, 7, 8, 9 or 10 classes, but not to anytwo
or more classes. In both these examples, occurrence of an event prevents the occurrence of
other events. Suchevents are called mutuallyexclusive events.
Two or moreevents ofan experiment, where occurence of anevent prevents occurences
ofallother events, are called Mutually Exclusive Events. We will discuss this in more detail
later inthe chapter.
Pierre Simon Laplace
(1749 – 1827)

Probability 313
13.4.1 FINDING PROBABILITY
How do we find theprobabilityofevents that are equallylikely?Weconsider the tossing
of acoin as an event associated with experiments wherethe equallylikelyassumptionholds. In
order to proceed, we recallthat there are two possible outcomes each time. Thisset ofoutcomes
is called the sample space. We can say that the sample space of one toss is {H, T}. For the
experiment of drawing out a ball from a bag containing red, blue, yellow and white ball, the
sample space is {R, B,Y, W}. What is the sample space when a dice is thrown?
DO THIS
Think of 5 situations with equallylikelyevents and find the sample space.
Let usnow try to findthe probabilityof equallylikelyevents that are mutuallyexclusive.
Example-1. Find the probability of getting a head when a coin is tossed once.Also find the
probabilityofgetting a tail.
Solution : Inthe experiment oftossing a coin once, the number ofpossible outcomesis two -
Head (H) andTail(T). Let E be the event 'getting a head'. The number of outcomes favourable
to E, (i.e., ofgetting a head) is 1. Therefore,
P(E) = P (head) =
Number of outcomes favourable to E
Number of all possible outcomes
=
1
2
Similarly, ifFis the event 'gettinga tail', then
P(F) = P(tail) =
1
2
(Guess why?)
Example-2. A bag contains a red ball, a blue balland an yellow ball, allthe balls being of the
same size. Manasa takes out a ballfromthe bag without looking into it. What is the probability
that she takes a (i) yellow ball? (ii) red ball?(iii) blue ball?
Solution : Manasatakes out a ballfromthe bag without lookinginto it. So, it is equallylikely
that she takes out anyone ofthem.
Let Ybe the event 'the balltaken out is yellow', B be the event 'the balltakenout is blue',
and R be the event 'the balltaken out is red'.
Now, the number ofpossible outcomes= 3.
(i) The number ofoutcomes favourable to the eventY= 1.
So, P(Y) =
1
3
. Similarly, P(R) =
1
3
and P(B) =
1
3

Class-X Mathematics
314
Remarks
1. An event having onlyone outcome inan experiment is called anelementary event. In
Example 1, both the events E and F are elementaryevents. Similarly, in Example2, all
the three events,Y, B and R are elementary events.
2. In Example 1, we note that : P(E) + P(F) = 1
In Example 2, we note that : P(Y) + P(R) + P(B) = 1.
Ifwe find the probabilityofall the elementaryeventsand add them, wewould get the
totalas 1.
3. In events like a throwing a dice, probabilityof getting less than3 and of gettinga 3 or
more than three are not elementary events of the possible outcomes. In tossing two
coins {HH}, {HT}, {TH} and {TT} are elementaryevents.
Example-3. Suppose we throw a die once. (i) What is the probability of getting a number
greater than 4?(ii) What is the probabilityofgetting a number less than or equal to 4?
Solution : (i) Inrolling an unbaised dice
Sample space S = {1, 2, 3, 4, 5, 6}
No. ofoutcomes n(S) = 6
Favourable outcomes for E = {5, 6}
number greater than 4
No. offavourable outcomes n(E) = 2
Probability P(E) =
2
6
=
1
3
(ii) Let F be the event 'getting a number less than or equalto 4'.
Sample space S = {1, 2, 3, 4, 5, 6}
No. ofoutcomes n(S) = 6
Favourable outcomes for F = {1, 2, 3, 4}
number less or equalto 4
No. offavourable outcomes n(F) = 4
Probability P(F) =
4
6
=
2
3

Probability 315
Note : Are theevents E and F in the above example elementaryevents?
No, theyare not elementary events. The event E has 2 outcomes and the event F has 4
outcomes.
13.4.2 COMPLEMENTARY EVENTS AND PROBABILITY
Intheprevioussectionwe readabout elementaryevents.Theninexample-3, wecalculated
probabilityofevents whichare not elementary. We saw,
P(E) + P(F) =
1
3
+
2
3
= 1
Here F is the same as 'not E' because there are onlytwo events.
We denote the event 'not E' by E . This is called the complement event ofevent E.
So, P(E) + P(not E) = 1
i.e., P(E) + P( E ) = 1, which givesus P( E ) = 1 - P(E).
In general, it is true that for an event E, P( ) = 1 – P(E)
DO THIS
(i) Is getting a head complementary to getting a tail? Give reasons.
(ii) In case of a die is getting a 1 complementary to events getting 2, 3, 4, 5, 6? Give reasons
for your answer.
(iii) Write of any five pair of events that are complementary.
13.4.3 IMPOSSIBLE AND CERTAIN EVENTS
Consider the following about the throws of a dice with sides marked as 1, 2, 3, 4, 5, 6.
(i) What is the probability of getting a number 7 in a single throw of a dice?
We know that there are only six possible outcomes in a single throw of this die. These
outcomes are 1, 2, 3, 4, 5 and 6. Since no face of the die is marked 7, there is no outcome
favourable to 7, i.e., the number of such outcomes is zero. In other words, getting 7 in a single
throw of a die, is impossible.
So P(getting 7) =
0
6
= 0
That is, the probability of an event which is impossible to occur is 0. Such an event is
called an impossible event.
(ii) What is the probability of getting 6 or a number less than 6 in a single throw of a dice?

Class-X Mathematics
316
Since every face of a die is marked with 6 or a number less than 6, it is sure that we will
always get one of these when the dice is thrown once. So, the number of favourable outcomes is
the same as the number of all possible outcomes, which is 6.
Therefore, P(E) = P(getting 6 or a number less than 6) =
6
6
= 1
So, the probability of an event which is sure (or certain) to occur is 1. Such an event is
called a sure event or a certain event.
Note : From the definition of probability P(E), we see that the numerator (number of outcomes
favourable to the event E) is always less than or equal to the denominator (the number of all
possible outcomes). Therefore, 0 £ P(E) £ 1.
TRY THIS
1. A child has a die whose six faces show the letters A, B, C, D, E and F. The die is thrown
once. What is the probability of getting (i) A? (ii) D?
2. Which of the following cannot be the probability of an event?
(a) 2.3 (b) -1.5 (c) 15% (D) 0.7
THINK AND DISCUSS
1. Why is tossing a coin considered to be a fair way of deciding which team should get the
ball at the beginning of any game?
2. Can
7
2
be the probability of an event? Explain.
3. Which of the following arguments are correct and which are not correct? Give reasons.
i) If two coins are tossed simultaneously there are three possible outcomes - two heads,
two tails or one of each. Therefore, for each of these outcomes, the probability is
1
3
.
ii)If a die is thrown, there are two possible outcomes - an odd number or an even number.
Therefore, the probability of getting an odd number is
1
2
.
13.5 DECK OF CARDS AND PROBABILITY
Have you seen a deck of playing cards?
A deck of playing cards consists of 52 cards which are divided into 4 suits of 13 cards
each. They are black spades (ª), red hearts (©), red diamonds (¨) and black clubs (§).

Probability 317
The cards in each suit are Ace, King, Queen, Jack, 10, 9, 8, 7, 6, 5, 4, 3 and 2. Kings,
Queens and Jacks are called face cards. Manygames are played with this deck of cards, some
games areplayed with part
of the deck and some with
two decks even. The study
ofprobabilityhasa lot to do
withcardanddicegamesas
it helps players to estimate
possibilitiesandpredicthow
the cards could be
distributed among players.
Example-4. One card isdrawnfroma well-shuffleddeck of52 cards. Calculate the probability
that the card will (i) be an ace, (ii) not be an ace.
Solution :Well-shuffling ensuresequallylikelyoutcomes.
(i) There are 4 aces in a deck.
Let E be the event 'the card is an ace'.
The number ofoutcomes favourable to E = 4
The number of possible outcomes =52 (Why?)
Therefore, P(E) =
4 1
=
52 13
(ii) Let F be the event 'card drawn is not an ace'.
The number of outcomes favourable to the event F = 52 - 4 = 48 (Why?)
The number ofpossible outcomes = 52
Therefore, P(F) =
48 12
=
52 13
Alternate Method : Note that F is nothing but E .
Therefore, we can also calculate P(F) as follows:
P (F) = P( E ) = 1 - P(E) = 1 -
1 12
=
13 13
TRY THIS
You have a singledeck ofwellshuffled cards. Then,
1. What is the probabilitythat the card drawn willbe a queen?

Class-X Mathematics
318
2. What is theprobability that it is a face card?
3. What is the probabilityit is a spade?
4. What is the probability that is the face card ofspades?
5. What is the probabilityit is not a face card?
13.6 USE OF PROBABILITY
Let us look at some more occasions where probability may be useful. We know that in
sports some countries are strong and others are not so strong. We also know that when two
players areplaying it is not that theywinequaltimes. The probabilityofwinning ofthe player or
team that wins more often is more than the probability of the other player or team. We also
discuss and keep track ofbirthdays. Sometimes it happens that people we knowhave the same
birthdays. Canwe find out whetherthisisa commonevent or would it onlyhappenoccasionally.
Classicalprobabilityhelpsus do this.
Example-5. Sangeeta and Reshma, play a tennis match. It is known that the probability of
Sangeetawinningthematchis 0.62. What is the probabilityofReshmawinning the match?
Solution : Let S and R denote the events that Sangeeta wins the match and Reshma wins the
match, respectively.
The probabilityofSangeeta's winning chances = P(S)= 0.62 (given)
The probabilityof Reshma's winning chances = P(R) = 1 - P(S)
= 1 -0.62 = 0.38 [R and S are complementary]
Example-6. Sarada and Hamida are friends. What is the probability that both will have (i)
different birthdays? (ii) the same birthday?(ignoring a leap year).
Solution : Out ofthe two friends, one girl, say, Sarada's birthdaycan be anydayof the year.
Now, Hamida's birthdaycan also be anydayof365 days inthe year. We assume that these 365
outcomes areequallylikely.
(i) If Hamida's birthdayis different fromSarada's, the number of favourable outcomes for her
birthday is 365 - 1 = 364
So, P (Hamida's birthdayis different fromSarada's birthday) =
364
365
(ii) P(Sarada and Hamida have the same birthday) = 1 - P (both have different birthdays)
= 1-
364
365
[ Using P( E ) = 1 - P(E)] =
1
365

Probability 319
Example-7. There are 40 students in Class X of a school of whom 25 are girls and 15 are
boys. The class teacher has to select one student as a class representative. She writes the name
of each student on separate cards, the cards being identical. Thenshe puts cards in a box and
stirs themthoroughly. She then draws one card fromthe box. What is the probabilitythat the
name written onthe card is the name of(i) a girl? (ii) a boy?
Solution : There are 40 students, and onlyone name card has to be chosen.
The number ofallpossible outcomes is 40
(i) The number ofoutcomes favourable for a card with the name ofa girl= 25 (Why?)
P (card with name ofa girl) = P(Girl) =
25 5
=
40 8
(ii) The number of outcomes favourable for a card with the name of a boy= 15 (Why?)
Therefore, P(card withname ofa boy) = P(Boy) =
15 3
=
40 8
or P(Boy) = 1 - P(not Boy) = 1 - P(Girl) = 1 -
5 3
=
8 8
EXERCISE - 13.1
1. Complete thefollowingstatements:
(i) Probability ofan event E + Probabilityofthe event 'not E' = ______________
(ii) The probabilityofan event that cannot happen is__________.
Such an event is called __________
(iii) The probabilityofan event that is certain to happen is __________.
Such an event is called______
(iv) Thesumofthe probabilitiesofallthe elementaryeventsofanexperiment is_________
(v) The probability ofan event is greater than or equalto __________ and less than or
equalto _______
2. Whichofthe following experimentshave equallylikelyoutcomes?Explain.
(i)Adriver attempts to start a car. The car starts or does not start.
(ii)Aplayer attempts to shoot a basketball. She/he shoots or misses the shot.
(iii)Atrialis made to answer a true-false question.The answer is right or wrong.
(iv)Ababyis born. It is a boyor a girl.

Class-X Mathematics
320
3. If P(E) = 0.05, what is the probabilityof 'not E'?
4. Abag containslemonflavoured candies only. Malinitakesout one candywithout looking
into the bag. What is the probabilitythat she takes out
(i) anorange flavoured candy? (ii) a lemon flavoured candy?
5. Rahimremovesallthe hearts fromthe cards. What isthe probabilityof
i. Getting anace fromthe remaining pack.
ii. Getting adiamonds.
iii. Getting acard that is not a heart.
iv. Getting theAce ofhearts.
6. It is giventhat in a group of 3 students, the probabilityof2 students not having the same
birthdayis 0.992. What is the probabilitythat the 2 students havethe same birthday?
7. A die is rolled once. Find the probabilityofgetting
(i) a prime number (ii) a number lying between 2 and 6 (iii) anodd number.
8. What is the probabilityofselecting a red king froma deck ofcards?
9. Make 5 more problems ofthiskind using dice, cardsor birthdays and discusswithfriends
and teacher about their solutions.
13.7 MORE APPLICATIONS OF PROBABILITY
We have seen some example of use of probability. Think about the contents and ways
probabilityhas beenused in these. Wehave seenagainthat probabilityof complementaryevents
add to 1. Can you identifyin the examples and exercises given above, and those that follow,
complementary events and elementary events? Discuss with teachers and friends. Let us see
more uses.
Example-8. A box contains 3 blue, 2 white, and 4 red marbles. If a marble is selected at
randomfromthe box, what isthe probabilitythat it willbe
(i) white? (ii) blue? (iii) red?
Solution : Saying that a marble is drawnat randommeans allthe marbles are equallylikely to be
drawn.
The number of possible outcomes = 3 +2 + 4 = 9 (Why?)
Let Wdenotethe event 'the marbleis white', Bdenotetheevent 'themarbleisblue' and R
denote the event 'marble is red'.
(i) The number ofoutcomes favourable to the event W= 2

Probability 321
So, P(W) =
2
9
Similarly, (ii) P(B) =
3 1
=
9 3
and (iii) P(R) =
4
9
Note that P(W) + P(B) + P(R) = 1.
Example-9. Harpreet tosses two different coins simultaneously(say, one is ofD1 andother of
D2). What is the probabilitythat she gets at least one head?
Solution : We write H for 'head'and T for 'tail'. When two coins are tossed simultaneously, the
possible outcomes are (H, H), (H, T), (T, H), (T, T), whichare allequallylikely. Here (H, H)
means heads on the first coin (sayon D1) and also heads on the second coin (D2). Similarly (H,
T) means heads up on the first coin and tailup on the second coinand so on.
The outcomes favourable to the event E, 'at least one head' are(H, H), (H, T) and (T, H).
So, the number ofoutcomes favourable to E is 3.
P(E) =
3
4
[Since the totalpossible outcomes= 4]
i.e., the probabilitythat Harpreet gets at least one head is
3
4
Check This
Didyouobservethat inalltheexamplesdiscussedsofar, thenumber ofpossibleoutcomes
in each experiment was finite? Ifnot, check it now.
Therearemanyexperimentsinwhichtheoutcomeisnumber betweentwo givennumbers,
or inwhichthe outcomeis everypoint withina circleorrectangle,etc. Canyoucount the number
ofallpossibleoutcomes insuch cases?As you know, thisis not possible sincethere are infinitely
manynumbersbetweentwo givennumbers,orthereare infinitelymanypointswithinacircle.So,
the definition of theoreticalprobability which you have learnt so far cannot be applied in the
present form.
What is thewayout?To answerthis, let us considerthe following example:
Example-10. In a musicalchair game, the personplaying the music has been advised to stop
playing themusicat anytime within2minutesafter shestartsplaying. What isthe probabilitythat
the music willstop withinthe firsthalf-minuteafter starting?

Class-X Mathematics
322
Solution : Herethe possible outcomes are allthe numbers between0 and 2. This is the portion
ofthe number line from0 to 2
Let E bethe event that 'themusic is stopped withinthe first half-minute'.
The outcomes favourable to E are points onthe number line from0 to
1
2
The distance from 0 to 2 is 2, while the distance from0 to
1
2
is
1
2
Since allthe outcomes are equallylikely, we can argue that, ofthe totaldistance is 2 and
the distance favourable to the event E is
1
2
So, P(E) =
Distance favourable to the event E
Total distance in which outcomes can lie
=
1
1
2
2 4
=
Wenowtryto extendthisideaforfindingtheprobabilityastheratio ofthefavourablearea
to thetotal area.
Example-11. A missing helicopter is
reported to havecrashed somewhere in the
rectangular region as shown in the figure.
What istheprobabilitythat it crashedinside
the lake showninthe figure?
Solution : The helicopter is equally likely
to crash anywhere in the region. Area of
the entire regionwhere the helicopter can
crash = (4.5 × 9) km2
= 40.5 km2
Area of the lake = (2.5 × 3) km2
= 7.5 km2
Therefore, P (helicopter crashed in the lake) =
7.5 5
=
40.5 27
= 0.185
Example-12. A cartonconsists of100 shirtsofwhich 88 aregood, 8 have minordefects and 4
have major defects. Jhony, a trader, will only accept the shirts which are good, but Sujatha,
another trader, will only reject the shirts which have major defects. One shirt is selected at
randomfromthe carton. What isthe probabilitythat
(i) it isacceptable to Jhony? (ii) it isacceptable to Sujatha?
0 1
2
1 2
6 .
km
Lake
9 .
km
4.5
.
km
2
.
km

Probability 323
Solution : Oneshirt is selected at randomfrom the carton of100 shirts. Therefore, there are
100 equallylikelyoutcomes.
(i) The number ofoutcomes favourable (i.e., acceptable) to Jhony= 88 (Why?)
Therefore, P (shirt is acceptable to Jhony) =
88
100
= 0.88
(ii) The number ofoutcomes favourable to Sujatha = 88 + 8 = 96 (Why?)
So, P (shirt is acceptable to Sujatha) =
96
100
= 0.96
Example-13. Two dice, one red and one yellow, are thrownat the same time.Write down all
the possible outcomes. What istheprobabilitythat the sumofthe two numbersappearing onthe
top of the dice is (i) 8 (ii) 13 (iii) less than or equal to 12?
Solution : When the red dice shows '1', the yellow dice could show anyone ofthe numbers 1,
2, 3, 4, 5, 6. The same is true when the red dice shows '2', '3', '4', '5' or '6'. The possible
outcomesoftheexperiment areshown inthe figure; the first number in eachorderedpair isthe
number appearing on the red dice and the
second number is that on the white dice.
Note that the pair (1, 4) is different
from(4, 1). (Why?)
So, the numberofpossible outcomes
n(S) = 6 × 6 = 36.
(i) Theoutcomesfavourableto theevent
'the sum of the two numbers is 8' denoted
byE, are: (2, 6), (3, 5), (4, 4), (5, 3), (6, 2) (See figure)
i.e., thenumber ofoutcomes favourable to E is n(E) = 5.
Hence, P(E) =
n(E)
n(S)
=
5
36
(ii) As there is no outcome favourable to the event F, 'the sum of two numbers is 13',
So, P(F) =
0
36
= 0
(iii)As allthe outcomes are favourable to the event G, 'sumoftwo numbers is 12',
So, P(G) =
36
36
= 1
1, 1
1 2 3 4 5 6
1
2
3
4
5
6
2, 1
3, 1
4, 1
5, 1
6, 1
1, 2
2, 2
3, 2
4, 2
5, 2
6, 2
1, 3
2, 3
3, 3
4, 3
5, 3
6, 3
1, 4
2, 4
3, 4
4, 4
5, 4
6, 4
1, 5
2, 5
3, 5
4, 5
5, 5
6, 5
1, 6
2, 6
3, 6
4, 6
5, 6
6, 6

Class-X Mathematics
324
EXERCISE - 13.2
1. Abag contains 3 red balls and 5 black balls. Aball is selected at random from the bag.
What is the probabilitythat the ballselected is (i) red ?(ii) not red?
2. A boxcontains 5 red marbles,8 white marbles and 4 green marbles. One marble is taken
out ofthe boxat random. What is the probabilitythat the marble taken out willbe (i) red?
(ii) white ?(iii) not green?
3. A Kiddy bank contains hundred 50p coins, fifty D1 coins, twenty D2 coins and ten D5
coins. Ifit isequallylikelythat one ofthecoinswillfallout whenthe bank isturned upside
down, what isthe probabilitythat the coin(i) willbe a 50p coin?(ii) willnot be a D5coin?
4. Gopi buys a fish from a shop for his aquarium. The shopkeeper
takes out onefishat randomfroma tank containing 5male fishand
8 femalefish(Seefigure). Whatis theprobabilitythat thefishtaken
out is a malefish?
5. A game of chance consists ofspinning an arrow which comes to
rest pointing at one of the numbers 1, 2, 3, 4, 5, 6, 7, 8 (See
figure), and these are equally likely outcomes. What is the
probabilitythat it willpoint at
(i) 8 ? (ii) anodd number?
(iii) a number greater than 2? (iv) a number less than 9?
6. One cardis selected fromawell-shuffled deck of52cards. Find the probabilityofgetting
(i) akingofred colour (ii) a face card (iii) a red face card
(iv) the jack of hearts (v) a spade (vi) thequeen ofdiamonds
7. Five cards-the ten, jack, queen, king and ace of diamonds, are well-shuffled with their
face downwards. One card is selected at random.
(i) What is the probabilitythat the card is the queen?
(ii) Ifthe queen is selected andput aside (without replacement), what isthe probabilitythat
the second card selected is (a) an ace? (b) a queen?
8. 12 defectivepens are accidentallymixed with 132 good ones. It is not possibleto just look
at apenandtellwhetheror not itisdefective. One penistakenout at randomfromthislot.
Determine the probabilitythat the pentaken out is a good one.
9. A lot of20 bulbs contain 4 defective ones. One bulb is selected at randomfromthe lot.
What is the probabilitythat this bulb is defective?Suppose the bulb selected inprevious
case is not defective and is not replaced. Now one bulb is selected at random from the
rest. What isthe probabilitythat this bulbis not defective?
1
8
7
6
5 4
3
2

Probability 325
10. A box contains 90 discs which are numbered from 1 to 90. If one disc is selected at
randomfromthebox, find the probabilitythat it bears(i)a two-digit number (ii) a perfect
square number (iii) a number divisibleby5.
11. Suppose you drop a die at random on the rectangular
regionshowninfigure. What isthe probabilitythat it will
land inside the circle with diameter 1m?
12. A lot consists of144 ballpens ofwhich 20 are defective
and the others are good. The shopkeeper draws one
penatrandomandgivesittoSudha.What istheprobability
that (i) She will buyit?(ii) She will not buyit ?
13. Two dice are rolled simultaneously and counts are added (i) complete the table given
below:
Event : 'Sumon 2 dice' 2 3 4 5 6 7 8 9 10 11 12
Probability
1
36
5
36
1
36
(ii)Astudent argues that 'there are 11 possible outcomes 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 and
12. Therefore, eachofthemhas aprobability
1
11
. Doyouagreewiththisargument?Justify
your answer.
14. A game consists of tossing a one rupee coin 3 times and noting its outcome each time.
Deskhitha wins ifallthe tosses give the same result i.e.,three heads or three tails, and loses
otherwise. Calculate the probabilitythat she willlose the game.
15. Adiceisthrowntwice. What isthe probabilitythat (i) 5 willnot come up either time?(ii) 5
will come up at least once? [Hint : Throwing a dice twice and throwing two dice
simultaneouslyare treated as the same experiment].
OPTIONAL EXERCISE
1. Two customers Shyamand Ekta are visiting aparticular shop inthe sameweek (Tuesday
to Saturday). Eachis equallylikely to visit the shop on anyday as on another day. What is
the probabilitythat bothwillvisit the shop on (i) the same day? (ii) consecutive days? (iii)
different days?
2. A bag contains 5 red balls and some blue balls. If the probabilityofdrawing a blue ballis
double that ofa red ball, determine the number ofblue balls in the bag.
3. A box contains 12 balls out ofwhich x are black. If one ballis drawn at randomfromthe
box, what is the probabilitythat it willbe ablack ball?If6 more black balls areput inthe
box, the probabilityofdrawing a black ballis now double ofwhat it was before. Find x.
3 .
m
2
.
m

Class-X Mathematics
326
4. A jar contains 24 marbles, some are green and others are blue. If a marble is drawn at
randomfromthe jar, the probabilitythat it isgreenis 2
3
. Findthe number ofblue marbles
in the jar.
In thischapter, you have studiedthe following points:
1. We havedealt withexperimentalprobabilityandtheoreticalprobability.
2. The theoretical(classical) probabilityofan event E, written as P(E), is defined as
P (E) =
Number of outcomes favourable to E
Total number of all possible outcomes of the experiment
where we assume that the outcomes ofthe experiment are equallylikely.
3. The probabilityofa sure event (or certainevent) is 1.
4. The probabilityofan impossible event is 0.
5. The probabilityofan event E is a number P(E) suchthat 0 £ P (E) £ 1
6. Aneventhavingonlyoneoutcomeiscalledanelementaryevent.Thesumoftheprobabilities
ofallthe elementary events ofanexperiment is 1.
7. For any event E, P (E) + P ( E ) = 1, where E stands for 'not E'. E and E are called
complementaryevents.
8. Some more terms used inthe chapter are given below:
Random experiment : For random experiments, the results are known well in advance,
but the result of the specific performance cannot be predicted.
Equally likely events : Two or more events are said to be equally likely if each one of
them has an equal chance of occurrance.
Mutually Exclusive events : Two or more events are mutually exclusive if the occurrence of
each event prevents the every other event.
Exhaustive events : Two or more events are said to be exhaustive, if the union of
their outcomes is the entire sample space.
Complementary events : Two events are said to be complementary, if they are mutually
exclusive and also exhaustive. (OR) Two events are said to be
complementary if occurrence of an event prevents the occurrence
of the other and the union of their outcomes is the entire sample
space.
Sure events : An event which will definitely occur is called a sure event. The
collection of all outcomes of a sure event is theentire sample space.
Impossible event : An event which cannot occur on any account is called an
impossible event.
Suggested Projects
Comparing classicalprobabilitywith experimentalprobability.
l Findingprobabilityforvarioussituationslikewhenadiceisrolled100timestheprobability
ofgetting (i) even, (ii) odd, (iii) prime etc.

14.1 INTRODUCTION
Ganesh recorded the marks of 26 children in his class in the mathematics Summative
Assessment -I in the register as follows:
Arjun 76 Narayana 12
Kamini 82 Suresh 24
Shafik 64 Durga 39
Keshav 53 Shiva 41
Lata 90 Raheem 69
Rajender 27 Radha 73
Ramu 34 Kartik 94
Sudha 74 Joseph 89
Krishna 76 Ikram 64
Somu 65 Laxmi 46
Gouri 47 Sita 19
Upendra 54 Rehana 53
Ramaiah 36 Anitha 69
Whether the recordeddata is organised properlyor not? Why?
Histeacheraskedhimtoreport onhowhisclassstudentshaveperformedinmathematics
in their SummativeAssessment - I .
Statistics
14

Class-X Mathematics
328
Ganesh prepared the following table to understand the performance ofhis class:
Marks Number ofchildren
0 - 33 4
34 - 50 6
51 - 75 10
76 - 100 6
Is the data given in the above table grouped or ungrouped?
He showed this table to his teacher and the teacher appreciated himfor organising the
data to be understood easily. Wecansee that most childrenhave got marks between 51-75. Do
you think that Ganeshshould haveused smaller range?Whyor whynot?
Inthepreviousclass,youhadlearntabout thedifferencebetweengroupedandungrouped
data aswellashow to present thisdata inthe formoftables.You had also learnt to calculate the
mean value for ungrouped data. Let us recallthis learning and then learnto calculate the mean,
median and modefor grouped data.
14.2 MEAN OF UNGROUPED DATA
We know that the mean (or average)ofobservations is theratio ofsum ofthe values of
alltheobservationsdivided bythe totalnumber ofobservations. Let x1
, x2
,. . ., xn
beobservations
with respective frequencies f1
, f2
, . . ., fn
. This means that observation x1
occurs f1
times, x2
occurs f2
times, and so on.
Now, the sum of the values of all the observations = f1
x1
+ f2
x2
+ . . . + fn
xn
, and the
number of observations = f1
+ f2
+ . . . + fn
.
So, the mean x ofthe data isgiven by
1 1 2 2
1 2
... ... ...
... ... ...
n n
n
f x f x f x
x
f f f
∗ ∗ ∗
<
∗ ∗ ∗
Recallthat we can write this in short, using the Greek letter å (read as sigma) which
means summationi.e., i i
i
f x
x
f
å
<
å
Example-1. The marks obtained in mathematics by30 students of Class X ofa certain school
are givenin table the below. Find the mean ofthe marks obtained bythe students.
Marks obtained (xi ) 10 20 36 40 50 56 60 70 72 80 88 92 95
Number of student ( fi) 1 1 3 4 3 2 4 4 1 1 2 3 1

Statistics 329
Solution : Let usre-organize this data and find the sumofallobservations.
Marks Number of i i
f x
obtained (xi ) students ( fi)
10 1 10
20 1 20
36 3 108
40 4 160
50 3 150
56 2 112
60 4 240
70 4 280
72 1 72
80 1 80
88 2 176
92 3 276
95 1 95
Total i
f
å =30 i i
f x
å = 1779
So,
1779
59.3
30
i i
i
f x
x
f
å
< < <
å
Therefore, the meanmarks are 59.3.
In most ofour reallife situations,data is usuallyso large that to make a meaningfulstudy,
it needs to be condensed as a grouped data. So, we need to convert ungrouped data into
grouped data and derive some method to find its mean.
Let us convert the ungrouped data ofExample 1 into grouped data byforming class-
intervals ofwidth, say 15. Remember that while allocating frequencies to each class-interval,
studentswhose score isequalto in any upperclass-boundary would be considered inthe next
class, e.g., 4 students who have obtained 40 marks would be considered in the class-interval
40-55 and not in 25-40. With this convention in our mind, let us form a grouped frequency
distributiontable.
Class interval 10-25 25-40 40-55 55-70 70-85 85-100
Number ofstudents 2 3 7 6 6 6

Class-X Mathematics
330
Now, for eachclass-interval, we require apoint which would serve as the representative
of the whole class. It is assumed that the frequency of each class-interval is centred
around its mid-point. So, the mid-point of each class can be chosen to represent the
observations falling inthat class and iscalled the class mark. Recallthat we find the class mark by
finding the average of the upper and lower limit ofthe class.
Upper class limit + Lower class limit
Class mark =
2
For the class 10 -25, the class mark is
10 25
2
∗
=17.5. Similarly, we can find the class
marks ofthe remaining class intervals. We put theminthe table.These class marks serve as our
xi
’s. We cannow proceed to compute the mean in thesame manner as inthe previous example.
Class Number of Class fi
xi
interval students ( fi
) Marks (xi
)
10-25 2 17.5 35.0
25-40 3 32.5 97.5
40-55 7 47.5 332.5
55-70 6 62.5 375.0
70-85 6 77.5 465.0
85-100 6 92.5 555.0
Total i
f
å =30 i i
f x
å =1860.0
The sumofthe values in the last columngives us i i
f x
å . So, the mean x ofthe given
data is given by
1860
62
30
i i
i
f x
x
f
å
< < <
å
This new method of finding the mean is knownas the Direct Method.
We observe that in the above cases we are using the same data and employing the same
formula for calculating the meanbut the results obtained are different. In example (1), 59.3 is the
exact mean and 62 is the approximatemean. Can you think whythis is so?

Statistics 331
THINK AND DISCUSS
1. The meanvaluecanbe calculated fromboth ungrouped and grouped data. Whichone
do you think is more accurate? Why?
2. When is it more convenient to use groupeddata for analysis?
Sometimes when thenumericalvalues ofxi and fi are large, findingthe product ofxi and fi
becomestediousandtime consuming. So,forsuchsituations, letus thinkofamethodofreducing
thesecalculations.
We can do nothing with the fi's, but wecanchange each xi to a smaller number so that our
calculations become easy. How do we do this?How about subtracting a fixed number fromeach
ofthese xi's? Let us trythis method for the data inexample 1.
The first step is to choose oneamong the xi's as the assumed mean, and denote it by'a'.
Also, to further reduce our calculation work, we maytake 'a'to be that xi whichliesinthe centre
of x1, x2, ..., xn. So, we can choose a = 47.5 or a = 62.5. Let us choose a = 47.5.
The second step is to find the deviation of ‘a’ from each of the xi's, which we denote
as di
i.e., di = xi – a = xi – 47.5
The third step is to find the product of di withthe corresponding fi, and take the sumof all
the fi di’s. These calculations are shown intable given below-
Class Number of Class 47.5
i i
d x
< , i i
f d
interval students ( fi) Marks (xi ) i i
d x a
< ,
10-25 2 17.5 -30 -60
25-40 3 32.5 -15 -45
40-55 7 47.5 (a) 0 0
55-70 6 62.5 15 90
70-85 6 77.5 30 180
85-100 6 92.5 45 270
Total i
f
å =30 i i
f d
å = 435
So, fromthe above table, the mean ofthe deviations, i i
i
f d
d
f
å
<
å
Now, let us find the relation between d and x .

Class-X Mathematics
332
Since, inobtaining di we subtracted ‘a’ fromeach xi so, inorder to get the mean x we
need to add ‘a’ to d . Thiscan be explained mathematicallyas:
Mean ofdeviations, i i
i
f d
d
f
å
<
å
So,
( )
i i
i
f x a
d
f
å ,
<
å
= i i i
i i
f x f a
f f
å å
,
å å
= i
i
f
x a
f
å
,
å
d = x a
,
Therefore
i i
i
f d
x a
f
å
< ∗
å
Substituting the values of a, i i
f d
å and i
f
å fromthe table, we get
435
47.5 47.5 14.5 62
30
x < ∗ < ∗ <
Therefore, the mean ofthe marks obtained by the students is 62.
The method discussed above is called the Assumed Mean Method.
ACTIVITY
Consider the data given in example 1 and calculate the arithmetic mean bydeviation
method bytakingsuccessive values ofxi i.e., 17.5, 32.5, ... as assumedmeans. Now discuss
thefollowing:
1. Are the values ofarithmetic meanin allthe abovecases equal?
2. If we take theactualmean as the assumed mean, how much will i i
f d
å be?
3. Reason about taking anymid-value (class mark) as assumed mean?
Observe that in the table given below the values in Column 4 are all multiples of 15. If
we divide all the values of Column 4 by 15, we would get smaller numbers which we then
multiplywith fi. (Here, 15 is the classsize ofeachclass interval.)
So, let
i
i
x a
u
h
,
< , where a is the assumed mean and h is the class size.

Statistics 333
Now, we calculate ui in this way and continue as before (i. e., find fi ui and then
å fi ui). Taking h = 15 [generallysize of the classis taken as h but it need not be sizeofthe class
always].
Let i i
i
f u
u
f
å
<
å
.
Class Number of Class i i
d x a
< ,
i
i
x a
u
h
,
< i i
f u
interval students ( fi) Marks (xi)
10-25 2 17.5 -30 -2 -4
25-40 3 32.5 -15 -1 -3
40-55 7 47.5 0 0 0
55-70 6 62.5 15 1 6
70-85 6 77.5 30 2 12
85-100 6 92.5 45 3 18
Total i
f
å =30 i i
f u
å =29
Here again, let us find the relation between u and x .
We have
i
i
x a
u
h
,
<
So i i
i
f u
u
f
å
<
å
( )
i
i
i
x a
f
h
u
f
,
å
<
å
=
1 i i i
i i
f x f a
h f f
é ù
å å
ê ú
,
ê ú
å å
ë û
=
1
( )
x a
h
,
hu x a
< ,
x a hu
< ∗
Therefore, i i
i
f u
x a h
f
é ù
å
ê ú
< ∗
ê ú
å
ë û
.

Class-X Mathematics
334
or
i i
i
f u
x a h
f
æ ö
å ÷
ç ÷
< ∗ ´
ç ÷
ç ÷
ç å
è ø
Substituting the values ofa, i i
f u
å hand i
f
å from the table, we get
29
47.5 15
30
x < ∗ ´
= 47.5 + 14.5 = 62
So, the mean marks obtained by a student are 62.
The method discussed above is called the Step-deviation method.
We note that:
l The step-deviation method will be convenient to apply if all the di’s have a common
factor.
l The mean obtained byallthe threemethods is same.
l The assumedmeanmethod and step-deviationmethod are just simplified forms ofthe
direct method.
l The formula x a hu
< ∗ still holds if a and h are not as given above, but are anynon-
zero numberssuchthat
i
i
x a
u
h
,
< .
Let us applythese methods inmore examples.
Example-2. The tablebelow gives the percentagedistribution of female teachers in the primary
schools of rural areas of various states and union territories (U.T.) of India. Find the mean
percentage offemale teachers using allthe three methods.
Percentage of female teachers 15-25 25-35 35 – 45 45-55 55-65 65-75 75– 85
Number of States/U.T. 6 11 7 4 4 2 1
Source : Seventh All India School Education Survey conducted by NCERT
Solution : Let us find the class marks xi of eachclass, and arrange themin a table.
Here, we take a = 50 and h = 10.
Then di = xi – 50 and ui =
50
10
i
x ,
.

Statistics 335
Now find di and ui and write themin the table
Percentage Number of i
x i
d < i
u < i i
f x i i
f d i i
f u
of female States/U.T. 50
i
x ,
50
10
i
x ,
teachers C.I fi
15– 25 6 20 -30 -3 120 -180 -18
25 – 35 11 30 -20 -2 330 -220 -22
35 – 45 7 40 -10 -1 280 -70 -7
45 – 55 4 50 0 0 200 0 0
55 – 65 4 60 10 1 240 40 4
65 – 75 2 70 20 2 140 40 4
75 – 85 1 80 30 3 80 30 3
Total 35 1390 -360 -36
Fromthe abovetable, we obtain 35, 1390, 360, 36
i i i i i i i
f f x f d f u
å < å < å <, å <, .
Using the direct method,
1390
39.71
35
i i
i
f x
x
f
å
< < <
å
.
Using the assumed mean method
360
50 50 10.29 39.71
35
i i
i
f d
x a
f
å ,
< ∗ < ∗ < , <
å
.
Using the step-deviation method
36
50 10 39.71
35
i i
i
f u
x a h
f
æ ö
å ,
÷
ç ÷
< ∗ ´ < ∗ ´ <
ç ÷
ç ÷
ç å
è ø
.
Therefore, themean percentage offemale teachers in the primaryschools ofruralareas is 39.71.
THINK AND DISCUSS
1. Is the result obtained byall the three methods same?
2. Ifxi and fi are sufficientlysmall, then which method is an appropriate choice?
3. Ifxi and fi are numericallylarge numbers, thenwhich methods are appropriate to use?
Even ifthe class sizes are unequal, and xi are large numerically, we can stillapply the
step-deviation method bytaking h to be a suitable divisor ofallthe di’s.
Example-3. The below distributionshowsthe numberofwickets taken bybowlersin one-day
cricket matches. Find the mean number ofwickets bychoosing a suitable method. What does
themeansignify?
Number ofwickets 20 - 60 60 - 100 100 - 150 150 - 250 250 - 350 350 – 450
Number of bowlers 7 5 16 12 2 3

Class-X Mathematics
336
Solution : Here, the class size varies, and the xi's are large. Let us stillapplythe step deviation
method with a = 200 and h = 20. Then, we obtain the data as given inthe table.
Number of Number of i
x i
d < i
u <
i
x a
h
,
i i
f u
wickets bowlers ( fi ) i
x a
, (h = 20)
20 – 60 7 40 -160 -8 -56
60 – 100 5 80 -120 -6 -30
100 – 150 16 125 -75 -3.75 -60
150 – 250 12 200 (a) 0 0 0
250 – 350 2 300 100 5 10
350 – 450 3 400 200 10 30
Total 45 -106
So
106
200 20 200 47.11 152.89
45
i i
i
f u
x a h
f
æ ö
å ,
÷
ç ÷
< ∗ ´ < ∗ ´ < , <
ç ÷
ç ÷
ç å
è ø
Thus, the average number of wickets taken bythese 45bowlers in one-daycricket is 152.89.
Classroom Project :
1. Collect themarks obtained byallthe students ofyour class inMathematics inthe recent
examination conducted in your school. Forma grouped frequencydistribution ofthe
data obtained. Do the same regarding other subjects and compare. Find the mean in
each case using a method you find appropriate.
2. Collect the dailymaximumtemperatures recorded for a period of 30 days in your city.
Present this data as a grouped frequency table. Find the mean of the data using an
appropriate method.
3. Measure the heights of all the students of your class and form a grouped frequency
distribution table ofthis data. Find the mean ofthe data using anappropriate method.
EXERCISE - 14.1
1. A surveywas conducted bya group ofstudents asa part oftheirenvironment awareness
programme, in whichtheycollected the following data regarding the number ofplants in
20 housesin a locality. Find the mean number ofplants per house.
Number of plants 0 - 2 2 - 4 4 – 6 6 - 8 8 - 10 10 - 12 12 – 14
Number of houses 1 2 1 5 6 2 3

Statistics 337
2. Consider the following distribution ofdailywagesof50 workersofa factory.
Dailywagesin Rupees 200 - 250 250 - 300 300 - 350 350 - 400 400– 450
Number of workers 12 14 8 6 10
Findthemeandailywagesoftheworkersofthefactorybyusinganappropriatemethod.
3. The followingdistributionshowsthedailypocket allowance ofchildrenofalocality.The
mean pocket allowance is ` 18. Find the missing frequency f.
Dailypocket 11 - 13 13 - 15 15 - 17 17 - 19 19 - 21 21 - 23 23 - 25
allowance(inRupees)
Number ofchildren 7 6 9 13 f 5 4
4. Thirtywomenwere examined in a hospitalbya doctor and their heart beatsper minute
were recorded and summarised as shown. Find the mean heart beats per minute for
these women, choosing a suitable method.
Numberofheart beats/minute 65-68 68-71 71-74 74-77 77-80 80-83 83-86
Number ofwomen 2 4 3 8 7 4 2
5. In a retail market, fruit vendors were selling oranges kept in packing baskets. These
baskets contained varying number of oranges. The following was the distribution of
oranges.
Number of oranges 10-14 15–19 20-24 25-29 30–34
Number of baskets 15 110 135 115 25
Find the mean number of oranges kept in each basket. Which method of finding the
mean did you choose?
6. The table belowshows the dailyexpenditure on food of25 householdsin a locality.
Dailyexpenditure(inRupees) 100-150 150-200 200-250 250-300 300-350
Number ofhouse holds 4 5 12 2 2
Find the meandailyexpenditure on food bya suitable method.
7. To find out the concentration ofSO2 inthe air (inpartsper million, i.e., ppm), the data
was collected for 30 localities ina certaincity and is presented below:
Concentration of SO2 in ppm 0.00-0.04 0.04-0.08 0.08-0.12 0.12-0.16 0.16-0.20 0.20-0.24
Frequency 4 9 9 2 4 2
Find the meanconcentration ofSO2 inthe air.

Class-X Mathematics
338
8. A class teacher has the following attendance record of 40 students of a class for the
whole term. Find the mean number ofdays a student was present out of56 days in the
term.
Number of days 35-38 38-41 41-44 44-47 47-50 50-53 53-56
Number of students 1 3 4 4 7 10 11
9. The following table gives the literacy rate (in percentage) of 35 cities. Find the mean
literacyrate.
Literacyrate in % 45–55 55-65 65-75 75-85 85-95
Number ofcities 3 10 11 8 3
14.3 MODE
Amode isthat value amongthe observations whichoccursmost frequently.
Before learning, how to calculate the mode ofgrouped data, let us first recallhow we
found the modefor ungrouped data throughthe following example.
Example-4. The wickets taken bya bowler in 10 cricket matches are asfollows: 2, 6, 4, 5, 0,
2, 1, 3, 2, 3. Find the mode of the data.
Solution : Let us arrange the observations in order i.e., 0, 1, 2, 2, 2, 3, 3, 4, 5, 6
Clearly, 2 is the number of wickets taken by the bowler in the maximum number of matches
(i.e., 3 times). So, the mode of this data is 2.
DO THIS
1. Find the modeofthe following data.
a) 5, 6, 9, 10, 6, 12, 3, 6, 11, 10, 4, 6, 7.
b) 20, 3, 7, 13, 3, 4, 6, 7, 19, 15, 7, 18, 3.
c) 2, 2, 2, 3, 3, 3, 4, 4, 4, 5, 5, 5, 6, 6, 6.
2. Is the mode always at the centre of the data?
3. Does the mode change, if another observation is added to the data in example?
Comment.
4. If the maximum value of an observation in the data in Example 4 is changed to 8,
would the mode ofthe data be affected? Comment.

Statistics 339
Ina groupedfrequencydistribution, it isnot possible to determine themode bylooking
at the frequencies. Here, we can only locate a class with the maximum frequency, called the
modalclass. The mode is a value inside the modalclass, and is givenbythe formula.
Mode 1 0
1 0 2
2
f f
l h
f f f
æ ö
, ÷
ç ÷
< ∗ ´
ç ÷
ç ÷
ç , ,
è ø
where, l = lower boundary ofthe modal class,
h = size ofthe modalclass interval,
f1 = frequencyof the modalclass,
f0 = frequencyof the class preceding the modal class,
f2 = frequencyofthe class succeeding the modalclass.
Let us consider the following examples to illustrate the use ofthis formula.
Example-5. Asurveyconducted on 20 householdsin a localitybya group ofstudents resulted
in the followingfrequencytable for the number offamilymembersina household.
Familysize 1-3 3-5 5-7 7-9 9-11
Number offamilies 7 8 2 2 1
Find the mode ofthis data.
Solution : Here themaximumclass frequencyis8, and the class corresponding to this frequency
is 3-5. So, the modal class is 3-5.
Now,
modal class = 3-5, boundary limit (l) of modalclass= 3, class size (h) = 2
frequencyofthe modalclass (f1) = 8,
frequency of class preceding the modalclass (f0) = 7,
frequencyofclass succeeding the modal class (f2) = 2.
Now, let us substitute these values in the formula-
Mode 1 0
1 0 2
2
f f
l h
f f f
æ ö
, ÷
ç ÷
< ∗ ´
ç ÷
ç ÷
ç , ,
è ø
8 7 2
3 2 3 3.286
2 8 7 2 7
æ ö
, ÷
ç
< ∗ ´ < ∗ <
÷
ç ÷
ç
è ø
´ , ,
Therefore, the mode ofthe data above is 3.286.
Example-6. The marks distributionof30 students ina mathematics examinationaregiveninthe
adjacent table. Find the mode ofthis data.Also compare and interpret the mode and the mean.

Class-X Mathematics
340
Class interval Number of Class Marks (xi ) fixi
students ( fi)
10-25 2 17.5 35.0
25-40 3 32.5 97.5
40-55 7 47.5 332.5
55-70 6 62.5 375.0
70-85 6 77.5 465.0
85-100 6 92.5 555.0
Total i
f
å = 30 i i
f x
å = 1860.0
Solution : Since the maximumnumber ofstudents(i.e.,7) have got marksinthe interval, 40-65
the modal class is 40 - 55.
The lower boundary( l ) ofthe modal class= 40,
the class size ( h) = 15,
the frequency ofmodal class ( f1 ) = 7,
the frequencyofthe class preceding the modalclass ( f0 ) = 3 and
the frequencyofthe class succeeding the modalclass ( f2 ) = 6.
Now, usingthe formula:
Mode 1 0
1 0 2
2
f f
l h
f f f
æ ö
, ÷
ç ÷
< ∗ ´
ç ÷
ç ÷
ç , ,
è ø
7 3
40 15 40 12 52
2 7 6 3
æ ö
, ÷
ç
< ∗ ´ < ∗ <
÷
ç ÷
ç
è ø
´ , ,
Interpretation : The mode marksis 52. Now, fromExample 1, we know that the mean marks
is 62. So, the maximumnumber ofstudents obtained 52 marks, while on anaverage a student
obtained 62 marks.
THINK AND DISCUSS
1. It depends uponthe demand ofthe situation whether we are interested in finding the
averagemarksobtainedbythestudentsorthemarksobtainedbymostofthestudents.
a. What do we find in the first situation?
b. What do we find inthe second situation?
2. Can mode be calculated for grouped data withunequalclass sizes?

Statistics 341
EXERCISE - 14.2
1. Thefollowingtableshowstheagesofthepatientsadmittedinahospitalonaparticularday:
Age (in years) 5-15 15-25 25-35 35-45 45-55 55-65
Number of patients 6 11 21 23 14 5
Find the mode and the mean ofthe data given above. Compare and interpret the two
measuresofcentraltendency.
2. The following data gives the information on the observed life times (inhours) of 225
electricalcomponents :
Lifetimes(in hours) 0 - 20 20 - 40 40 - 60 60 - 80 80 - 100 100 - 120
Frequency 10 35 52 61 38 29
Determine the modallifetimesofthe components.
3. The followingdata gives the distributionoftotalmonthlyhouseholdexpenditure of200
families ofGummadidala village. Find themodalmonthlyexpenditure ofthe families.
Also, find themeanmonthlyexpenditure :
Expenditure 1000- 1500- 2000- 2500- 3000- 3500- 4000- 4500-
(in rupees) 1500 2000 2500 3000 3500 4000 4500 5000
Number of families 24 40 33 28 30 22 16 7
4. The followingdistributiongivesthestate-wise, teacher-student ratio inhigher secondary
schools ofIndia. Find the mode and mean ofthis data. Interpret the two measures.
Number ofstudents 15-20 20-25 25-30 30-35 35-40 40-45 45-50 50-55
Number of States 3 8 9 10 3 0 0 2
5. The given distribution shows the number ofruns scored by some top batsmen of the
world in one-dayinternationalcricket matches.
Runs 3000- 4000- 5000- 6000- 7000- 8000- 9000- 10000-
4000 5000 6000 7000 8000 9000 10000 11000
Number of batsmen 4 18 9 7 6 3 1 1
Find the mode ofthe data.

Class-X Mathematics
342
6. A student noted the number of cars passing through aspot on a road for 100 periods,
each of3 minutes, and summarised this in the table givenbelow.
Number of cars 0 - 10 10 - 20 20 - 30 30 - 40 40 - 50 50 - 60 60 - 70 70 - 80
Frequency 7 14 13 12 20 11 15 8
Find the mode ofthe data.
14.4 MEDIAN OF GROUPED DATA
Median is a measure of central tendency which gives the value of the middle-most
observation inthe data. Recallthat for finding the medianofungrouped data, we first arrange the
data valuesor the observations inascending order.
Then, if n is odd, the median is the
1
2
th
n
æ ö
∗ ÷
ç ÷
ç ÷
ç
è ø
observationand
if n is even, thenthe median willbe the averageofthe
2
th
n
æ ö
÷
ç ÷
ç ÷
ç
è ø
and 1
2
th
n
æ ö
÷
ç ∗ ÷
ç ÷
ç
è ø
observations.
Suppose, we have to find the median of the following data, which is about the marks, out of
50 obtained by 100 students ina test :
Marks obtained 20 29 28 33 42 38 43 25
Numberof students 6 28 24 15 2 4 1 20
First, we arrange the marks inascending order and prepare a frequencytable 14.9 as follows :
Marks obtained Number ofstudents (frequency)
20 6
25 20
28 24
29 28
33 15
38 4
42 2
43 1
Total 100

Statistics 343
Here n = 100, which is even. The median will be the average of the
2
th
n
æ ö
÷
ç ÷
ç ÷
ç
è ø
and the 1
2
th
n
æ ö
÷
ç ∗ ÷
ç ÷
ç
è ø
observations, i.e., the 50
th
and 51
st
observations. To find the positionofthesemiddle values, we
construct cumulativefrequency.
Marks obtained Number of students Cumulativefrequency
20 6 6
upto 25 6 + 20 = 26 26
upto 28 26 + 24 = 50 50
upto 29 50 + 28 = 78 78
upto 33 78 + 15 = 93 93
upto 38 93 + 4 = 97 97
upto 42 97 + 2 = 99 99
upto 43 99 + 1 = 100 100
Now we add another column depicting this information to the frequencytable above andname it
as cumulative frequency column.
From thetable above, we seethat :
50
th
observation is 28 (Why?)
51
st
observationis 29
Median =
28 29
28.5
2
∗
<
Remark: Column1and column 3 inthe above table
are known as Cumulative Frequency Table. The
medianmarks28.5conveystheinformationthatabout
50% students obtained marks less than 28.5 and
another50%studentsobtainedmarksmorethan28.5.
Consider a grouped frequencydistribution of
marksobtained, outof100, by53students,inacertain
examination, as shownin adjacent table.
Marks Number of students
0-10 5
10-20 3
20-30 4
30-40 3
40-50 3
50-60 4
60-70 7
70-80 9
80-90 7
90-100 8

Class-X Mathematics
344
Fromthe table, tryto answer the following questions :
How many students have scored marks
less than 10? The answer is clearly5.
How manystudentshave scoredless than
20 marks? Observe that the number of
students who have scored less than 20
include the numberofstudentswho have
scored marks from 0-10 as well as the
number of students who have scored
marks from10-20. So, the total number
ofstudents withmarks less than20 is 5 +
3, i.e., 8. We say that the cumulative
frequency of the class 10-20 is 8. (As
shown intable 14.11)
Similarly, wecancompute the cumulative
frequencies ofthe other classes, i.e., the
number of students with marks less than
30, less than 40, ..., less than 100.
Thisdistributionis called the cumulativefrequencydistributionoftheless thantype. Here 10, 20,
30, ..., 100, are the upper boundaries of the respective class intervals.
Wecansimilarlymakethe tablefor
thenumberofstudentswithscores
more than or equal to 0 (this
number is same as sum of all the
frequencies), morethanabovesum
minus the frequency of the first
class interval, more than or equal
to 20 (this number is same as the
sum of all frequencies minus the
sumofthe frequencies ofthe first
two class intervals), and so on.
We observe that all 53 students
have scored marks more than or
equal to 0. Since there are 5
students scoring marks in the
interval0-10,thismeansthat there
Marks obtained Number of students
(Cumulativefrequency)
Less than 10 5
Less than 20 5 + 3 = 8
Less than 30 8 + 4 = 12
Less than 40 12 + 3 = 15
Less than 50 15 + 3 = 18
Less than 60 18 + 4 = 22
Less than 70 22 + 7 = 29
Less than 80 29 + 9 = 38
Less than 90 38 + 7 = 45
Less than 100 45 + 8 = 53
Marks obtained Number of students
(Cumulativefrequency)
More than or equalto 0 53
More than or equalto 10 53 - 5 = 48

Statistics 345
are 53-5 =48 students getting morethanor equalto 10 marks. Continuing inthe same manner,
we get the number ofstudents scoring 20 or above as 48-3 = 45, 30or above as 45-4= 41, and
so on, as shown in the table aside.
This tableabove is called a cumulative frequencydistribution ofthe more thantype. Here 0, 10,
20, ..., 90 give the lower boundaries ofthe respective class intervals.
Now, to find the medianofgrouped data, we can make use ofany ofthese cumulative frequency
distributions.
Now in a grouped data, we may not be able to find the middle observation by looking at the
cumulative frequencies as the middle observation will be some value in a class interval. It is,
therefore, necessaryto find the value inside a class that divides the whole distributioninto two
halves. But which class should this be?
To find thisclass, we find the cumulative frequencies of allthe classes and
2
n
.Wenow locate the
class whose cumulativefrequencyexceeds
2
n
for the first time. Thisiscalled the medianclass.
Marks Number ofstudents (f) Cumulative frequency(cf)
0-10 5 5
10-20 3 8
20-30 4 12
30-40 3 15
40-50 3 18
50-60 4 22
60-70 7 29
70-80 9 38
80-90 7 45
90-100 8 53
In the distribution above, n = 53. So
2
n
= 26.5. Now 60-70 is the class whose cumulative
frequency29 is greater than (and nearest to)
2
n
, i.e., 26.5.
Therefore, 60-70 is the median class.

Class-X Mathematics
346
After finding themedian class, we use the following formula for calculating the median.
Median 2
n
cf
l h
f
æ ö
÷
ç , ÷
ç ÷
ç ÷
ç
< ∗ ´
÷
ç ÷
ç ÷
ç ÷
÷
ç
è ø
where l = lower boundary ofmedian clas,
n = number of observations,
cf = cumulativefrequencyofclass preceding the median class,
f = frequencyof median class,
h = class size (size ofthe median class).
Substituting the values
2
n
=26.5, l = 60, cf = 22, f = 7, h = 10
in the formula above, we get
Median
26.5 22
60 10
7
é ù
,
ê ú
< ∗ ´
ê ú
ë û
45
60
7
< ∗
66.4
<
So, about half the students have scored marks less than 66.4, and the other half have scored
marks more than 66.4.
Example-7. Asurveyregarding the heights
(in cm) of 51 girls ofClass X ofa school was
conducted and data was obtained asshown in
table. Findtheir median.
Height (in cm) Number of girls
Less than 140 4
Less than 145 11
Less than 150 29
Less than 155 40
Less than 160 46
Less than 165 51

Statistics 347
Solution : To calculate the median
height, we need to find the class
intervals and their corresponding
frequencies. Thegiven distribution
being of the less than type, 140,
145, 150, . . ., 165 give the upper
limits of the corresponding class
intervals. So, the classes should be
below 140, 140 - 145, 145 - 150, .
. ., 160 - 165.
Observe that fromthe given distribution, wefind that there are 4 girlswithheight less than140,
i.e., thefrequencyofclass intervalbelow 140 is 4 . Now, there are 11 girls withheightslessthan
145and 4girlswithheight lessthan140.Therefore, thenumberofgirlswithheight inthe interval
140 – 145 is 11 – 4 = 7. Similarly, the frequencies can be calculated as shown in table.
Number ofobservations, n = 51
2
n
=
51
2
= 25.5
th
observation, which lies in the class 145 - 150.
145 – 150 is the median class
Then, l (the lower boundary) = 145,
cf (the cumulative frequencyofthe class preceding 145 – 150) = 11,
f (the frequencyof the median class145 – 150) = 18 and
h (the class size) = 5.
Using the formula, Median 2
n
cf
l h
f
æ ö
÷
ç , ÷
ç ÷
ç
è ø
< ∗ ´
∋ (
25.5 11
145 5
18
,
< ∗ ´
72.5
145
18
< ∗ = 149.03.
Class intervals Frequency Cumulative
frequency
Below 140 4 4
140-145 7 11
145-150 18 29
150-155 11 40
155-160 6 46
160-165 5 51

Class-X Mathematics
348
So, the median height ofthe girls is 149.03 cm. This means that the height ofabout 50% ofthe
girls islessthanthis height,and that ofother50% is greater thanthis height.
Example-8. The median of the following data is 525. Find the values of x and y, if the total
frequencyis 100. Here, CI stands for class intervaland Fr for frequency.
CI 0-100 100- 200- 300- 400- 500- 600- 700- 800- 900-
200 300 400 500 600 700 800 900 1000
Fr 2 5 x 12 17 20 y 9 7 4
Class intervals Frequency Cumulativefrequency
0-100 2 2
100-200 5 7
200-300 x 7+x
300-400 12 19+x
400-500 17 36+x
500-600 20 56+x
600-700 y 56+x+y
700-800 9 65+x+y
800-900 7 72+x+y
900-1000 4 76+x+y
Solution :
It is given that n = 100
So, 76 + x + y = 100, i.e., x + y = 24 (1)
The medianis 525, which liesin the class 500 – 600
So, l = 500, f = 20, cf = 36 + x, h = 100

Statistics 349
Usingthe formula
Median
2
n
cf
l h
f
æ ö
÷
ç , ÷
ç ÷
ç
è ø
< ∗ ´
525
50 36
500 100
20
x
, ,
< ∗ ´
i.e., 525 – 500 = (14 – x) × 5
i.e., 25 = 70 – 5x
i.e., 5x = 70 – 25 = 45
So, x = 9
Therefore, from(1), we get 9 + y = 24
i.e., y = 15
Note :
The medianofgrouped data with unequalclass sizes canalso be calculated.
14.5 WHICH VALUE OF CENTRAL TENDENCY
Which measure would be best suited fora particularrequirement.
The meanis the most frequentlyused measure ofcentraltendency because it takes into account
allthe observations, and lies between theextremes, i.e., the largest and the smallest observations
of the entire data. It also enables us to compare two or more distributions. For example, by
comparingtheaverage(mean) resultsofstudentsofdifferent schoolsofaparticular examination,
we canconclude which schoolhas a better performance.
However, extreme values in the data affect the mean. For example, the mean ofclasses having
frequencies more or less the same is a good representative of the data. But, if one class has
frequency, say 2, and the five others have frequency 20, 25, 20, 21, 18, then the mean will
certainly not reflect the way the data behaves. So, in such cases, the mean is not a good
representative ofthe data.
In problemswhere individualobservations arenot important, especiallyextremevalues, and we
wish to find out a ‘typical’observation, the median is more appropriate, e.g., findingthe typical
productivityrate ofworkers, average wage ina country, etc. These aresituations where extreme
values may exist. So, rather than the mean, we take the median as a better measure of central
tendency.

Class-X Mathematics
350
Insituationswhichrequireestablishingthemost frequent valueormost popular item, themode is
the best choice, e.g., to find the most popular T.V.programmebeing watched, the consumer item
in greatest demand, the colour of the vehicle used bymost ofthe people, etc.
EXERCISE - 14.3
1. The following frequencydistributiongives themonthlyconsumptionofelectricityof68
consumersofalocality. Findthemedian,meanandmodeofthedataandcomparethem.
Monthly consumption 65-85 85-105 105-125 125-145 145-165 165-185 185-205
Number of consumers 4 5 13 20 14 8 4
2. Ifthe median of60 observations, given below is 28.5, find the values of x and y.
Class interval 0-10 10-20 20-30 30-40 40-50 50-60
Frequency 5 x 20 15 y 5
3. A life insurance agent found the following data about distribution ofages of100 policy
holders. Calculate the median age. [Policies are given only to persons having age 18
years onwards but less than 60 years.]
Age Below Below Below Below Below Below Below Below Below
(in years) 20 25 30 35 40 45 50 55 60
Number of 2 6 24 45 78 89 92 98 100
policy holders
4. The lengths of 40 leaves ofa plant are measured correct to the nearest millimetre, and
the data obtainedis represented in thefollowing table :
Length (in mm) 118-126 127-135 136-144 145-153 154-162 163-171 172-180
Number of leaves 3 5 9 12 5 4 2
Find the median length of the leaves. (Hint : The data needs to be converted to continuous
classes for finding the median, since the formula assumes continuous classes. The classes then
change to 117.5 - 126.5, 126.5 - 135.5, . . ., 171.5 - 180.5.)

Statistics 351
5. The followingtable gives the distributionofthe life-time of400 neonlamps
Lifetime 1500- 2000- 2500- 3000- 3500- 4000- 4500-
(inhours) 2000 2500 3000 3500 4000 4500 5000
Number of 14 56 60 86 74 62 48
lamps
Find the median life time ofa lamp.
6. 100surnameswererandomlypickedupfromalocaltelephonedirectoryandthefrequency
distributionofthenumber oflettersintheEnglishalphabet inthesurnameswasobtained
as follows
Number of letters 1-4 4-7 7-10 10-13 13-16 16-19
Number ofsurnames 6 30 40 16 4 4
Determine themedian number of lettersin the surnames. Find the mean number ofletters inthe
surnames?Also, find the modalsize ofthe surnames.
7. The distribution below gives the weights of 30 students of a class. Find the median
weight ofthestudents.
Weight(inkg) 40-45 45-50 50-55 55-60 60-65 65-70 70-75
Number ofstudents 2 3 8 6 6 3 2
14.6 GRAPHICAL REPRESENTATION OF CUMULATIVE FREQUENCY DISTRIBUTION
As we all know, pictures speak better than words. A graphical representation helps us in
understanding given data at a glance. In Class IX, we have represented the data through bar
graphs, histograms and frequency polygons. Let us now represent a cumulative frequency
distributiongraphically.
For example, let us consider thecumulative frequencydistributiongiveninexample.
Fordrawingogives,itshouldbeensuredthat theclassintervalsarecontinuous,becausecumulative
frequenciesarelinked withboundaries, but not withlimits.

Class-X Mathematics
352
Recall that the values 10, 20,
30, ..., 100 are the upper
boundaries of the respective
classintervals.To represent the
data graphically, we mark the
upper boundaries of the class
intervals onthe horizontalaxis
(X-axis) and their
corresponding cumulative frequencies on the verticalaxis (Y-axis), choosing a convenient scale.
Nowplot thepoints corresponding to theordered pairsgiven by(upper boundary, corresponding
cumulative frequency), i.e., (10, 5), (20, 8), (30,, 12), (40, 15), (50, 18), (60, 22), (70, 29),
(80, 38), (90, 45), (100, 53) on a graph paper and join them bya free hand smoothcurve. The
curve weget is called acumulative frequencycurve, or an ogive (of the less than type).
The term 'ogive is pronounced as 'ojeev' and is derived from the word ogee.An ogee is
a shape consisting of a concave arc flowing into a convex arc, so forming an S-shaped
curve with vertical ends. In architecture, the ogee shape is one ofthe characteristics of
the 14
th
and 15
th
century Gothic styles.
Again we consider the cumulative frequencydistributionand draw its ogive (ofthe more than
type).
Recall that, here 0, 10, 20, ...., 90
are the lower boundaries of the
respective class intervals 0-10, 10-
20, ....., 90-100. To represent 'the
more thantype'graphically, we plot
the lower boundariesonthe X-axis
and the corresponding cumulative
frequenciesontheY-axis. Then we
plot the points (lower boundaries,
corresponding cumulative
frequency), i.e., (0, 53), (10, 48),
(20, (45), (30, 41), (40, 38), (50,
35), (60, 31), (70, 24), (80, 15), (90, 8), ona graphpaper, and join thembya free hand smooth
curve. The curve we get is a cumulative frequency curve, or an ogive (of the more thantype).
Less
than
Cumulative
frequency
0
10
20
30
40
50
60
0 80 90 100
'Less than' ogive
Upper limits
'More than' ogive
0
10
20
30
40
50
60
0 10 20 30 40 50 60 70 80 90
Lower limits
Cumulative
frequency

Statistics 353
14.6.1 OBTAINING MEDIAN FROM A GIVEN CURVE:
Is it possible to obtain the median fromthese two cumulative frequencycurves . Let us see.
One obvious way is to locate on
53
26.5
2 2
n
< < on the y-axis. From this point, draw a line
parallel to the X-axis cutting the
curve at a point. From this point,
draw a perpendicular to the X-axis.
Foot of this perpendicular
determines the medianofthe data.
Another way of obtaining the
median :
Draw both ogives (i.e., of the less
than type and of the more than type)
onthesame axis. The two ogives willintersect eachother at a point. From this point, ifwe draw
a perpendicular on the x-axis, the point at which it cuts the x-axis givesus the median.
Example-9. The annual profits earned by 30 shops in Sangareddy locality give rise to the
followingdistribution:
Profit (in lakhs) Number ofshops (frequency)
Draw bothogives for the data above. Hence obtain themedianprofit.
Cumulative
frequency
Upper limits
Median (66.4)
0
10
20
30
40
50
60
0 10 20 30 40 50 60 70 80 90 100
Limits
Cumulative
frequency
Mediam (66.4)
Limits
0
10
20
30
40
50
60
0 10 20 30 40 50 60 70 80 90 100

Class-X Mathematics
354
Solution : We first drawthe coordinate axes, with lower limits ofthe profit along the horizontal
axis, and the cumulative frequencyalongthe verticalaxes. Then, we plot the points(5, 30), (10,
28), (15, 16), (20, 14), (25, 10), (30, 7) and (35, 3). We join these points with asmooth curve
to get the more than ovive, as shown inthe figure below-
Now, let us obtain the classes, their frequencies and the cumulative frequencyfrom the table
above.
Classes 5-10 10-15 15-20 20-25 25-30 30-35 35-40
Number ofshops 2 12 2 4 3 4 3
Cumulativefrequency 2 14 16 20 23 27 30
Using these values, we plot the points (10, 2), (15, 14), (20, 16), (25, 20), (30, 23), (35, 27),
(40, 30) on the same axes as in last figure to get the less than ogive, as shownin figure below.
The abcissa oftheir point of intersection is nearly 17.5, whichis the median. This can also be
verified byusing the formula. Hence, the median profit (in lakhs) is ` 17.5.
More
than
Cumulative
frequency
Lower limits of profit (in lakhs Rs.)
0
5
10
15
20
25
30
35
0 5 10 15 20 25 30 35

Statistics 355
EXERCISE - 14.4
1. The following distributiongives the dailyincome of50 workersofa factory.
Dailyincome (inRupees) 250-300 300-350 350-400 400-450 450-500
Number of workers 12 14 8 6 10
Convert thedistributionaboveto alessthantypecumulativefrequencydistribution,anddrawits
ogive.
2. During the medicalcheck-up of 35 students ofa class, their weights were recorded as
follows:
Weight(inkg) Number of students
Less than 38 0
Less than 40 3
Less than 42 5
Less than 44 9
Less than 46 14
Less than 48 28
Less than 50 32
Less than 52 35
Draw a less than type ogive for the givendata. Hence obtain the median weight fromthe graph
and verifythe result byusing the formula.
3. Thefollowingtablegivesproductionyieldper hectareofwheatof100farmsofavillage.
Productionyield 50-55 55-60 60-65 65-70 70-75 75-80
(Qui/Hec)
Number offarmers 2 8 12 24 38 16
Change thedistribution to a morethantypedistribution, and draw its ogive.
0
5
10
15
20
25
30
35
0 5 10 15 20 25 30 35 40
Median (17.5)
Profit (in lakhs Rs.)
Cumulative
frequency
Limits

Class-X Mathematics
356
Inthischapter, you have studied thefollowing points:
1. The meanfor grouped data is calculated by:
(i) The direct method : i i
i
f x
x
f
å
<
å
(ii) The assumed meanmethod : i i
i
f d
x a
f
å
< ∗
å
(iii) The step deviationmethod : i i
i
f u
x a h
f
æ ö
å ÷
ç ÷
< ∗ ´
ç ÷
ç ÷
ç å
è ø
2. The modefor grouped data canbe found byusingthe formula :
Mode 1 0
1 0 2
2
f f
l h
f f f
æ ö
, ÷
ç ÷
< ∗ ´
ç ÷
ç ÷
ç , ,
è ø
where, symbols havetheir usualmeaning.
3. The medianfor grouped data isformed byusing the formula :
Median 2
n
cf
l h
f
æ ö
÷
ç , ÷
ç ÷
ç ÷
ç
< ∗ ´
÷
ç ÷
ç ÷
ç ÷
÷
ç
è ø
, where symbols have their usualmeanings.
4. Inorder to find median, class intervalsshould be continuous.
5. Representing a cumulativefrequencydistributiongraphicallyas a cumulative frequency
curve, or an ogive ofthe lessthan type and ofthemore than type.
6. While drawing ogives, boundaries are takenon X-axis and cumulative frequencies are
taken onY-axis.
7. Scale onboth the axes maynot be equal.
8. The medianofgrouped data canbe obtained graphicallyas the x-coordinate ofthe point
ofintersectionofthe two ogivesfor this data.
Suggested Projects
Finding mean - median - mode.
l Applicationsofdailylife situation.
l Collecting informationfromavailable sources.
l Drawing graphs for mean, median and mode for the above collected data.

Mathematical Modelling
A.I.1 INTRODUCTION
On 25th
February2013, the ISRO launcherPSLV C20, placed the satellite SARALinto
orbit. The satelliteweighs 407 kg. It is at analtitudeof781 kmandits orbit is inclinedat anangle
of98.5º.
On reading the above information, wemaywonder:
(i) How did the scientists calculate thealtitude as 781km. Did theygo to spaceand measure
it?
(ii) How did theyconclude that the angle oforbit is 98.5ºwithout actuallymeasuring?
Somemoreexamples are there inour dailylife where we wonder how the scientists and
mathematicians could possiblyhave estimated these results. Observe these examples:
(i) The temperature at the surface of the sunis about 6,000ºC.
(ii) The human heart pumps 5 to 6 liters ofblood in the bodyeveryminute.
(iii) We know that the distance between the sun and the earth is 1,49,000 km.
Intheaboveexamples, weknowthat no onewentto thesuntomeasure thetemperature
or the distance fromearth. Nor can we take the heart out ofthe bodyand measure the blood it
pumps. Thewaywe answertheseand other similarquestionsisthroughmathematical modelling.
Mathematical modelling isused not onlybyscientists but also byus. For example, we
might want to know howmuchmoneywewillget afterone year ifwe invest D100 at 10% simple
interest. Or we might want to knowhow many litres ofpaint is needed to whitewash a room.
Eventhese problems are solved bymathematicalmodelling.
THINK AND DISCUSS
Discusswith your friends some more examples in reallife wherewe cannot directly
measure and must use mathematicalmodelling .

Class-X Mathematics
358
A.I.2 MATHEMATICAL MODELS
Do youremember the formula to calculate the area ofa triangle?
Area of Triangle =
1
2
´ base ´ height.
Similarly,simpleinterest calculationusestheformula
100
PTR
I < .Thisformulaorequation
is a relation between the Interest (I); Principle (P); Time (T); and Rate ofInterest (R). These
formulae are examplesofmathematicalmodels.
Some more examplesfor mathematicalmodels.
(i) Speed (S)
tan ( )
( )
Dis ce d
time t
<
(ii) Incompound interest sum(A) = 1
100
n
r
P
æ ö
÷
ç ∗ ÷
ç ÷
ç
è ø
Where P = Principle
r = rate of interest
n = no. of times to be calculated interest.
So, Mathematical model is nothing but a mathematical description or relation that
describes some real life situation.
DO THIS
Write some more mathematicalmodels whichyou have learnt inprevious classes.
A.I.3 MATHEMATICAL MODELLING
We often face problems in our dayto daylife. To solve them, we tryto write it as an
equivalent mathematicalproblemand find its solution. Next we interpret thesolutionand check
to what extent thesolutionisvalid. Thisprocess ofconstructinga mathematicalmodeland using
it to find the answer isknownasmathematicalmodelling.

Mathematical Modelling 359
Now we have to observe somemore examples related to mathematicalmodelling.
Example-1. Vani wants to buy a TV that costs D19,000 but she has only D15,000. So she
decides to invest her money at 8% simple interest per year.After how many years willshe be
able to buythe TV?
Step 1 : (Understanding the problem): In this stage, we define the realproblem. Here, we are
given the principal, the rate of simple interest andwe want to find out the number ofyears after
which the amount willbecome Rs. 19000.
Step2:(Mathematicaldescriptionandformulation) Inthisstep,wedescribe,inmathematical
terms, the different aspects ofthe problem. We define variables, writeequations or inequalities
and gather data ifrequired.
Here, we usethe formula for simple interest which is
100
PTR
I < (Model)
where P = Principle, T = number ofyears, R = rate of interest, I = Interest
We need to find time =
100I
T
RP
<
Step 3: (Solving the mathematical problem) In this step, we solve the problem using the
formula whichwe have developed instep 2.
We know that Vani already has D15,000 which is the principal, P
The final amount is D19000 so she needs an extra (19000-15000) = D4000. This will
come from the interest, I.
P = D15,000, Rate = 8%, then I = 4000;
100 4 0
T
´
<
0 0
150 0 0
40 0
8
<
´ 12 0
4 1
T 3 3
12 3
< < years
or Step4:(Interpretingthesolution): Thesolutionobtained inthepreviousstepis interpreted
here.
Here T = 3
1
3
. This means three and one third ofayear or three years and 4 months.
So, Vanican buy a TV after 3 year 4 months

Class-X Mathematics
360
Step5 : (Validating the model): We can’t always accept a modelthat gives us ananswer that
does not matchthe reality. The processofchecking and modifying themathematicalmodel, if
necessary, is validation.
Inthegivenexample, weareassumingthat therateofinterestwillnot change. Iftherate
changes thenour model
100
PTR
willnot work. We are also assuming that the price ofthe washing
machine willremain Rs. 19,000.
Let us takeanother example.
Example-2. InLokeshwaramHighschool, 50childreninthe10thclassandtheir Mathsteacher
want to go on tour fromLokeshwaramto Hyderabad byvehicles.A jeep canhold six persons
not including the driver. How manyjeeps theyneed to hire?
Step 1 : We want to find the number of jeeps needed to carry51 persons, given that eachjeep
can seat 6 persons besides the driver.
Step 2 : Number ofvehicles = (Number of persons) / (Persons that can be seated in one jeep)
Step 3 : Number of vehicles = 51/6 = 8.5
Step 4 : Interpretation
We know that it is not possible to have 8.5vehicles. So, the numberofvehicles needed
has to be the nearest whole number which is 9.
Number ofvehicles need is 9.
Step 5 : Validation
While modelling, we have assumed that lean and fat children occupysame space.
DO THIS
1. Take any word problem from your textbook, make a mathematical model for the
chosen problemand solve it.
2. Make a mathematicalmodelfor theproblemgiven below and solve it.
Suppose a car starts from a place A and travels at a speed of 40 Km/h towards
another place B.At the same time another car starts from B and travelstowardsAat
a speed of30 Km/h. Ifthe distance betweenAand B is 100 km;after how much time
willthat carsmeet?

So far, we have made mathematicalmodels for simple word problems. Let us take a real
life exampleand model it.
Example-3. Inthe year2000, 191 member countriesofthe U.N.signed adeclarationto promote
gender equality. One indicator for deciding whether this goalhas been achieved is the ratio of
girls to boys in primary, secondaryeducation. India also signed the declaration. The data for the
percentage ofgirls in India who are enrolled in primaryschools is given inTable A.I.1.
Table A.I.1
Year Enrolment (in %)
1991 – 92 41.9
1992 – 93 42.6
1993 – 94 42.7
1994 – 95 42.9
1995 – 96 43.1
1996 – 97 43.2
1997 -98 43.5
1998 – 99 43.5
1999 – 2000 43.6
2000 – 01 43.7
2001 - 02 44.1
Usingthisdata,mathematicallydescribetherateat whichtheproportionofgirlsenrolled
in primaryschools grew.Also, estimate the year bywhichthe enrolment ofgirlswillreach50%.
Solution :
Step 1 : Formulation Let us first convert the probleminto amathematical problem.
TableA.I.1 gives the enrolment for the years 1991 – 92, 1992- 93 etc. Since the students
join at the begining ofanacademic year, we cantake the years as 1991, 1992 etc. Let us assume
that thepercentageofgirlswho joinprimaryschoolswillcontinuetogrowat thesamerateasthe
rateinTableA.I.1. So, thenumberofyears isimportant, notthespecificyears. (Togivea similar
situation, when we find thesimple interest for say, ` 15000 at the rate 8% for three years, it does
not matter whether the three – year period is from1999 to 2002 orfrom 2001 to 2004. What is
important is the interest rate inthe years being considered)

Class-X Mathematics
362
Here also, we willsee howthe enrolment grows after 1991 bycomparing thenumber of
years that has passed after 1991 and the enrolment. Let us take 1991 asthe 0th
year, and write
1 for 1992 since 1 year has passed in 1992 after 1991. Similarly we will write 2 for 1993, 3 for
1994 etc. So, TableA.I.1 willnow look like asTableA.I.2
Table A.I.2
Year Enrolment (in%)
0 41.9
1 42.6
2 42.7
3 42.9
4 43.1
5 43.2
6 43.5
7 43.5
8 43.6
9 43.7
10 44.1
The increase inenrolment isgiveninthefollowing tableA.I.3.
Table A.I.3
Year Enrolment (in%) Increase
0 41.9 0
1 42.6 0.7
2 42.7 0.1
3 42.9 0.2
4 43.1 0.2
5 43.2 0.1
6 43.5 0.3
7 43.5 0
8 43.6 0.1
9 43.7 0.1
10 44.1 0.4

At the end of the first year period from 1991 to 1992, the enrolment has increased by
0.7% from 41.9% to 42.6%. At the end of the second year, this has increased by 0.1% from
42.6% to 42.7%. From the table above, we cannot find a definite relationship between the
number of yearsand percentage. But theincrease is fairlysteady. Onlyin the first year and in the
10th
year there is a jump. The mean ofthese values is
0.7 0.1 0.2 0.2 0.1 0.3 0 0.1 0.1 0.4
10
∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗
= 0.22 .... (1)
Let us assume that the enrolment steadilyincreases at the rate of 0.22 percent.
Step 2 : (Mathematical Description)
We have assumed that the enrolment increases steadily at the rate of0.22% per year.
So, the Enrolment Percentage (EP) inthe first year = 41.9 + 0.22
EP in the second year = 41.9 + 0.22 + 0.22 = 41.9 + 2 ´ 0.22
EP in the third year = 41.9 + 0.22 + 0.22 + 0.22 = 41.9 + 3 ´ 0.22
So, the enrolment percentage in the nth
year = 41.9 + 0.22n, for n ³1. .... (2)
Now, we also have to find the number ofyears bywhich the enrolment willreach 50%.
So, wehave to find the value ofn fromthis equation
50 = 41.9 + 0.22n
Step 3 : Solution : Solving (2) for n, we get
n =
50 41.9
0.22
,
=
8.1
0.22
= 36.8
Step 4 : (Interpretation) : Since the number ofyears is anintegral value, we willtake the next
higher integer, 37. So, the enrolment percentage willreach 50% in 1991 + 37= 2028.
Step 5 : (Validation) Since we are dealing with a real life situation, we have to see to what
extent this value matches the realsituation.
Let us check Formula (2) is inagreement with the reality. Let us find the values for the
years we alreadyknow, using Formula (2), and compare it with the known valuesbyfinding the
difference. Thevalues are giveninTableA.I.4.

Class-X Mathematics
364
TableA.I.4
Year Enrolment Values given by (2) Difference
(in %) (in %) (in %)
0 41.9 41.90 0
1 42.6 42.12 0.48
2 42.7 42.34 0.36
3 42.9 42.56 0.34
4 43.1 42.78 0.32
5 43.2 43.00 0.20
6 43.5 43.22 0.28
7 43.5 43.44 0.06
8 43.6 43.66 -0.06
9 43.7 43.88 -0.18
10 44.1 44.10 0.00
As you cansee, some ofthe values given byFormula (2) are less thanthe actualvalues
byabout 0.3% or even by0.5%. This can give rise to a difference ofabout 3 to 5years since the
increaseperyearisactually1% to2%.Wemaydecidethatthismuchofadifferenceisacceptable
and stop here. In this case, (2) isour mathematicalmodel.
Suppose we decide that this error is quite large, and we have to improve this model.
Then, we have to go back to Step 2, and change the equation. Let us do so.
Step 1 : Reformulation : We stillassume that the values increase steadily by0.22%, but we
willnow introducea correction factor to reduce the error, For this, we find the mean ofallthe
errors. Thisis
0 0.48 0.36 0.34 0.32 0.2 0.28 0.06 0.06 0.18 0
10
∗ ∗ ∗ ∗ ∗ ∗ ∗ , , ∗
= 0.18
We take the mean of the errors, and correct our formula bythis value.
Revised Mathematical Description : Let us now add the meanofthe errors to our formula
for enrolment percentage given in(2). So, our corrected formulais :

Enrolment percentage inthe nthyear
= 41.9 + 0.22n + 0.18 = 42.08 + 0.22n, for n ³1 ... (3)
We willalso modifyEquation (2) appropriately.The new equationforn is :
50 = 42.08 + 0.22n ... (4)
Altered Solution : Solving Equation(4) for n, we get
n =
50 42.08
0.22
,
=
7.92
0.22
= 36
Interpretation : Since n = 36, the enrolment ofgirls in primaryschools willreach 50% inthe
year 1991 + 36 = 2027.
Validation : Once again, let us compare the values got byusing Formula (4) withthe actual
values. TableA.I.5givesthe comparison.
TableA.I.5
Year Enrolment Values given Difference Values Difference
(in %) by (2) between given between
Values by (4) values
0 41.9 41.90 0 41.9 0
1 42.6 42.12 0.48 42.3 0.3
2 42.7 42.34 0.36 42.52 0.18
3 42.9 42.56 0.34 42.74 0.16
4 43.1 42.78 0.32 42.96 0.14
5 43.2 43.00 0.20 43.18 0.02
6 43.5 43.22 0.28 43.4 0.1
7 43.5 43.44 0.06 43.62 -0.12
8 43.6 43.66 -0.06 43.84 -0.24
9 43.7 43.88 -0.18 44.06 -0.36
10 44.1 44.10 0.00 44.28 -0.18
As you cansee, manyofthevalues that (4) gives are closer to the actualvalue than the
values that (2) gives. The mean ofthe errors is 0in this case.

Class-X Mathematics
366
A.I.4 ADVANTAGES OF MATHEMATICS MODELLING
1. The aimofmathematicalmodelling is to get some usefulinformationabout a realworld
problembyconvertingit into mathematicalproblem. This is especiallyusefulwhen it is
notpossibleorveryexpensiveto getinformationbyothermeanssuchasdirect observation
or byconductingexperiments.
For example, suppose we want to study the corrosive effect of the discharge of the
Mathura refineryon the TajMahal. We would not like to carryout experiments on the
TajMahaldirectlybecausethat woulddamage avaluablemonument.Heremathematical
modelling canbeofgreat use.
2. Forecasting isveryimportant inmanytypes oforganizations, sincepredictions offuture
events have to be incorporated into the decision– making process.
Forexample
(i) In marketing departments, reliable forecasts ofdemand help in planning ofthe
sale strategies
(ii) A school board needs to able to forecast the increase in the number of school
going children in various districts so as to decide where and when to start new
schools.
3. Often we need to estimate large valueslike trees in aforest; fishes in alake; estimation of
votes polled etc.
Some moreexamples where we usemathematical modelling are:
(i) Estimating future populationfor certain number ofyears
(ii) Predicting the arrivalof Monsoon
(iii) Estimating the literacyrate in coming years
(iv) Estimating number ofleaves in a tree
(v) Finding the depthofoceans
A.I.5 LIMITATIONS OF MATHEMATICAL MODELLING
Is mathematicalmodelling the answer to allour problems?
Certainlynot;it has it’s limitations.Thus,we should keep inmind that a modelis onlya
simplificationofarealworldproblem, andthetwo arenot same. It issomethinglikethedifference
between a map that gives the physicalfeatures of a country, and the country itself. We can find
the height ofa place above the sea levelfromthis map, but we cannot find the characteristics of
the people from it. So, we should use a model only for the purpose it is supposed to serve,
remembering allthefactors we have neglected while constructing it. We should applythe model
onlywithinthe limits where it isapplicable.

A.I.6 TO WHAT EXTENT WE SHOULD TRY TO IMPROVE OUR MODEL?
To improve a modelwe need to take into account severaladditionalfactors. When we
do thisweaddmorevariablestoour mathematicalequations. Theequationsbecomescomplicated
andthemodelisdifficult touse.Amodelmustbesimpleenoughtouseyet accurate;i.ethecloser
it is to realitythe betterthe modelis.
TRY THIS
Aproblemdating back to the early13th
century, posed by Leonardo Fibonacci, asks
how manyrabbitsyou would haveinone year ifyou started with just two and let allofthem
reproduce.Assume that a pair ofrabbits produces a pair of offspring each month and that
eachpair ofrabbits produces their first offspring at theageof2months. Monthbymonth, the
number of pairs ofrabbits is given by the sum ofthe rabbits in the two preceding months,
except for the 0th
and the 1st
months. The table below shows how the rabbit population
keeps increasing everymonth.
Month Pairs of Rabbits
0 1
1 1
2 2
3 3
4 5
5 8
6 13
7 21
8 34
9 55
10 89
11 144
12 233
13 377
14 610
15 987
16 1597
After one year, we have 233 rabbits.After just 16 months, we have nearly1600 pairs
of rabbits.
Clearlystate theproblemand the different stages ofmathematicalmodellinginthis situation.

Class-X Mathematics
368
We willfinishthis chapter bylooking at some interesting examples.
Example-4. (Rolling of a pairof dice) : Deekshitha andAshish are playing with dice. Then
Ashish said that, ifshe correctlyguess the sumofnumbers that show up on the dice, he would
give a prize for everyanswer to her. What numbers would be the best guess for Deekshitha.
Solution :
Step 1 (Understanding the problem) : You need to know a few numbers which have higher
chances ofshowing up.
Step 2(Mathematical description) : Inmathematicalterms, the problemtranslates to finding
out the probabilitiesofthe various possible sums ofnumbers that the dice could show.
We can model the situation very simply by representing a roll of the dice as a random
choice ofoneofthe following thirtysixpairsofnumbers.
(1, 1) (1, 2) (1, 3) (1, 4) (1, 5) (1, 6)
(2, 1) (2, 2) (2, 3) (2, 4) (2, 5) (2, 6)
(3, 1) (3, 2) (3, 3) (3, 4) (3, 5) (3, 6)
(4, 1) (4, 2) (4, 3) (4, 4) (4, 5) (4, 6)
(5, 1) (5, 2) (5, 3) (5, 4) (5, 5) (5, 6)
(6, 1) (6, 2) (6, 3) (6, 4) (6, 5) (6, 6)
The first number in each pair represents the number showing on the first die, and the
second number is the number showing onthe second die.
Step 3 (Solving the mathematical problem) : Summing the numbers in each pair above, we
find that possible sums are 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 and 12. We have to find the probability
for eachofthem, assuming all36 pairsare equallylikely.
We do this in the following table.
Sum 2 3 4 5 6 7 8 9 10 11 12
Probability
1
36
2
36
3
36
4
36
5
36
6
36
5
36
4
36
3
36
2
36
1
36
Observe that the chance of getting a sum of a seven is
1
6
, which is larger than the
chances ofgettingother numbersassums.

Step 4(Interpretingthe solution) : Sincetheprobabilityofgetting thesum7isthehighest, you
should repeatedlyguessthe number seven.
Step 5 (Validating the model) : Toss a pair of dice a large number of times and prepare a
relative frequencytable.Compare the relative frequencies withthe corresponding probabilities.
If these are not close, then possiblythe dice are biased. Then, we could obtaindata to evaluate
the number towardswhich the biasis.
Before goingto the next trythis exercise, weneed some background information.
Not having the moneyyou want when you need it, is a common experience for many
people. Whether it is having enoughmoneyfor buying essentials for dailyliving, or for buying
comforts, we always require money. To enable the customers with limited funds to purchase
goodslikescooters,refrigerators, televisions,cars,etc., aschemeknownasaninstalment scheme
(or plan) is introduced bytraders.
Sometimes a trader introduces an instalment scheme as a marketing strategy to allow
customers to purchase these articles. Under the instalment scheme, the customer is not required
to make fullpayment of the articleat the time ofbuying it. She/he is allowed to paya part ofit at
the time of purchase and the rest can be paid in instalments, which could be monthly, quarterly,
half-yearly, or even yearly. Of course, the buyer will have to pay more in the instalment plan,
becausethe seller isgoing to chargesome interest on account ofthe payment made at a later date
(called deferred payment).
There are some frequently used terms related to this concept. You maybe familiar with
them. For example, the cashprice of anarticle isthe amount which acustomer has to payas full
payment of the article at the time it is purchased. Cash down payment is the amount which a
customer has to payas part payment of the price ofan article at the time ofpurchase.
Now, tryto solve the problemgivenbelow byusing mathematicalmodelling.
TRY THIS
Ravi wants to buy a bicycle. He goes to the market and finds that the bicycle of his
choice costs `2,400. He has only `1,400 with him.To help, the shopkeepr offers to help
him. He says that Ravi can make a down payment of `1400 and pay the rest in monthly
instalments of `550 each. Ravican either take the shopkeepers offer or go to a bank and
takealoanat 12% perannumsimpleinterest. Fromthesetwo opportunitieswhichisthebest
one to Ravi. Help him.

Class-X Mathematics
370
EXERCISE - 1.1
1. (i) 90 (ii) 196 (iii) 127
EXERCISE - 1.2
1. (i) 22
≥ 5 ≥7 (ii) 22
≥ 3 ≥ 13 (iii) 32
≥ 52
≥17
(iv) 5 ≥ 7 ≥ 11
1 ≥13 (v) 17 ≥ 19 ≥ 23
2. (i) 420, 3 (ii) 1139, 1 (iii) 1800, 1
(iv) 216, 36 (v) 22338, 9
6. 6
EXERCISE - 1.3
1. (i) 0.375,Terminating (ii) 0.5725, Terminating (iii) 4.2, Terminating
(iv) 0.18, Non-terminating, repeating (v) 0.064,Terminating
2. (i) Terminating (ii) Non-terminating, repeating
(iii) Non-terminating, repeating (iv) Terminating
(v) Non-terminating (vi) Terminating
(vii) Non-terminating (viii) Terminating
(ix) Terminating (x) Non-terminating, repeating
3. (i) 0.52 (ii) 0.9375 (iii) 0.115 (iv) 32.08 (v) 1.3
4. (i) Rational, Prime factors of q will be either 2 or 5 or both only
(ii) Not rational
(iii) Rational, Prime factors of q willalso have a factor other than 2 or 5.
EXERCISE - 1.5
1. (i)
1
2
(ii)
1
4
(iii) -4 (iv) 0 (v)
1
2
(vi) 9
(vii) -2 (viii) 3 (ix) 12
Answers

Answers 371
2. (i) log 10, 1 (ii) log2
8, 3 (iii) log64
64, 1 (iv)
9
log
8
æ ö
÷
ç ÷
ç ÷
ç
è ø
(v) log 45
3. (i) x + y (ii) x + y ,1 (iii) x + y + 2 ( iv) 3x + 3y + 1
4. (i) 4 log 10 (ii) 7log 2 - 4 log 5 (iii) 2 log x + 3log y + 4 log z
(iv) 2 log p + 3log q - 4 log r (v)
3
2
log x - log y
6. 7 7.
1
3
8.
3
log
2
log6
æ ö
÷
ç ÷
ç ÷
ç
è ø
EXERCISE - 2.1
1. (i) Set (ii) Not set (iii) Not set
(iv) Set (v) Set
2. (i) Î (ii) Ï (iii) Ï (iv) Ï
(v) Î (vi) Î
3. (i) x Ï A (ii) d Î B (iii) 1 Î N (iv) 8 Ï P
4. (i) False (ii) False (iii) True (iv) False
5. (i) B = {1, 2, 3, 4, 5}
(ii) C = {17, 26, 35, 44, 53, 62, 71, 80}
(iii) D = {5, 3}
(iv) E = {B, E, T, R}
6. (i) A = {x : x is multiple of 3 & less than 13}
(ii) B = {x : x = 2a, a Î N, a < 6}
(iii) C = {x : x = 5a, a Î N, a < 5}
(iv) D = {x : x is square number and x < 10, x Î N}
7. (i) A = {51, 52, 53, .................., 98, 99}
(ii) B = {+2, ,2}
(iii) D = {L, O, Y, A}
8. (i) ,(c)

Class-X Mathematics
372
(ii) ,(a)
(iii) (d)
(iv) (b)
EXERCISE - 2.2
1. Yes,AÇ B &BÇ B are same
2. A Ç ε = ε
A Ç A = A
A
3. A ,B = {2, 4, 8, 10}
B ,A = {3, 9, 12, 15}
4. A È B = B
5. A Ç B = {even natural number}
{2, 4, 6, ............}
A Ç C = {odd naturalnumbers}
A Ç D = {4, 6, 8, 9, 10, 12, ................. 100}
B Ç C = f
B Ç D ={evennaturalnumber}
C Ç D = {3, 5, 7, 11, ..............}
6. (i) A , B = {3, 6, 9, 15, 18, 21}
(ii) A , C = {3, 9, 15, 18, 21}
(iii) A , D = {3, 6, 9, 12, 18, 21}
(iv) B , A= {4, 8, 16, 20}
(v) C , A = {2, 4, 8, 10, 14, 16}
(vi) D , A = {5, 10, 20}
(vii) B , C = {20}
(viii) B , D = {4, 8, 12, 16}
(ix) C , B = {2, 6, 10, 14)
(x) D , B = {5, 10, 15}
7. (i) False, because theyhave common element '3'

Answers 373
(ii) False, because the two sets have a common element 'a'
(iii) True, because no common elements for the sets.
(iv) True, because no common elements for the sets.
EXERCISE - 2.3
1. Yes, equalsets
2. (i) Equal(=) (ii) Not equal( ¹ ) (iii) Equal(=) (iv) Not equal( ¹ )
(v) Not equal( ¹ ) (vi) Not equal( ¹ ) (vii) Not equal( ¹ )
3. (i) A= B (ii) A ¹ B (iii) A= B (iv) A ¹ B
4. (i) {1, 2, 3, ..... 10} ¹ {2, 3, 4, ..... 9}
(ii) x = 2x + 1 means x is odd
(iii) x is multiple of15. So 5 does not exist
(iv) x is primenumber but 9 is not a prime number
5. (i) {p}, {q}, {p, q}, {ε}
(ii) {x}, {y}, {z}, {x, y}, {y, z}, {z, x}, {x, y, z}, ε
(iii) {a}, {b}, {c}, {d}, {a, b}, {b, c}, {c, d}, {a, c}, {a, d}, {b, d}, {a, b, c},
{b, c, d}, {a, b, d}, {a, c, d}, {a, b, c, d}, ε
(iv) ε, {1}, {4}, {9}, {16}, {1, 4}, {1, 9}, {1, 16}, {4, 9}, {4, 16}, {9, 16},
{1, 4, 9}, {1, 9, 16}, {4, 9, 16}, {1, 4, 16}, {1, 4, 9, 16}
(v) ε, {10}, {100}, {1000}, {10, 100}, {100, 1000}, {10, 1000},
{10, 100, 1000}
EXERCISE - 2.4
1. (i) Not empty (ii) Empty (iii) Empty
(iv) Empty (v) Not empty
2. (i) Finite (ii) Finite (iii) Finite
3. (i) Finite (ii) Infinite (iii) Infinite (iv) Infinite

Class-X Mathematics
374
EXERCISE - 3.1
1. (a) (i) ,6 (ii) 7 (iii) ,6
2. (i) False ( 2 is coefficient ofx2
not a degree)
(ii) False (Coefficient ofx2
is ,4)
(iii) True (For anyconstant term, degree is zero)
(iv) False (It is not a polynomialat all)
(v) False (Degree of a polynomial is not related with number ofterms)
3. p(1) = 0, p (,1) = ,2, p(0) = ,1, p(2) = 7, p(,2) = ,9
4. Yes, ,2 and ,2 are zeroes ofthe polynomial x4
-16
5. Yes, 3 and ,2 are zeroes of the polynomial x2
,x,6
EXERCISE - 3.2
1. (i) No zeroes (ii) 1 (iii) 3
(iv) 2 (v) 4 (vi) 3
2. (i) 0 (ii) ,2, ,3 (iii) ,2, ,3 (iv) ,2, 2, 4
° ,
3. (i) 4, ,3 (ii) 3, 3 (iii) No zeroes
(iv) ,4, 1 (v) ,1, 1
4. p
1
4
æ ö
÷
ç ÷
ç ÷
ç
è ø
= 0 and p (,1) = 0
EXERCISE - 3.3
1. (i) 4, ,2 (ii)
1 1
,
2 2
(iii)
3 1
,
2 3
,
(iv) 0, ,2 (v) 15, 15
, (vi)
4
1,
3
,
2. (i) 4x2
, x, 4 (ii) 2
3 2 1
x x
, ∗ (iii) 2
5
x ∗
(iv) 2
1
x x
, ∗ (v) 2
4 1
x x
∗ ∗ (vi) 2
4 1
x x
, ∗

Answers 375
3. (i) x2
, x, 2 (ii) 2
3
x , (iii) 2
4 3 1
x x
∗ ,
(iv) 2
4 8 3
x x
, ∗
4. ,1, +1 and 3 are zeros ofthe given polynomial.
EXERCISE - 3.4
1. (i) Quotient = x ,3 and remainder = 7x ,9
(ii) Quotient = x2
+ x ,3 and remainder = 8
(iii) Quotient = ,x2
,2 and remainder = ,5x + 10
2. (i) Yes (ii) Yes (iii) No
3. ,1, ,1
4. g(x) = x2
,x + 1
5. (i) p(x) = 2x2
,2x + 14, g(x) = 2, q(x) = x2
,x + 7, r(x) = 0
(ii) p(x) = x3
+ x2
+ x + 1, g(x) = x2
,1, q(x) = x + 1, r(x) = 2x + 2
(iii) p(x) = x3
+ 2x2
,x +2, g(x) = x2
,1, q(x) = x + 2, r(x) = 4
EXERCISE - 4.1
1. (a) Intersect at a point
(b) Coincident
(c) Parallel
2. (a) Consistent (b) Inconsistent (c) Consistent
(d) Consistent (e) Consistent (f) Inconsistent
(g) Inconsistent (h) Consistent (i) Inconsistent
3. Number ofpants = 1; Number ofshirts = 0
4. Number ofGirls = 7; Number of boys = 3
5. Cost of pencil = ` 3; Cost of pen = ` 5
6. Length = 20 m; Width = 16 m
7. (i) 6x , 5y ,10 = 0
(ii) 4x + 6y ,10 = 0
(iii) 6x + 9y ,24 = 0

Class-X Mathematics
376
8. Length= 40 units; Breadth = 30 units
9. Number ofstudents = 16; Number ofbenches = 5
EXERCISE - 4.2
1. Income of Ist
person = ` 18000; Income of IInd
person = ` 14000
2. 42 and 24
3. Angles are 81º and 99º
4. (i) Fixed charge = ` 40; Charge per km = ` 18 (ii) ` 490
5.
7
9
6. 60 km/h; 40 km/h.
7. 31º and 59º
8. 659 and 723
9. 40 mland 60 ml
10. ` 7200 and ` 4800
EXERCISE - 4.3
1. (i) (4, 5) (ii)
1 1
,
2 4
æ ö
, ÷
ç ÷
ç ÷
ç
è ø
(iii) (4, 9)
(iv) (1, 2) (v) (3, 2) (vi)
1 1
,
2 3
æ ö
÷
ç ÷
ç ÷
ç
è ø
(vii) (3, 2) (viii) (1, 1)
2. (i) Speed of boat = 8 km/h; Speed of streem= 3 km/h
(ii) Speed oftrain = 60 km/h; Speed of car = 80 km/h
(iii) Number of days byman = 18; Number of days by woman = 36
EXERCISE - 5.1
1. (i) Yes (ii) Yes (iii) No
(iv) Yes (v) Yes (vi) No
(vii) No (viii) Yes

Answers 377
2. (i) 2x2
+ x ,528 = 0 (x = Breadth)
(ii) x2
+ x ,306 = 0 (x = Smaller integer)
(iii) x2
+ 32x ,273 = 0 (x = Rohan's Age)
(iv) x2
, 8x , 1280 = 0 (x = Speed of the train)
EXERCISE - 5.2
1. (i) ,2; 5 (ii) ,2;
3
2
(iii) 2
, ;
5
2
,
(iv)
1 1
;
4 4
(v)
1 1
;
10 10
(vi) ,6; 2
(vii) 1,
2
3
(viii) ,1; 3 (ix) 7,
8
3
2. 13, 14
3. 17, 18; ,17, ,18
4. 5 cm, 12 cm
5. Number of articles = 6; Cost of each article = 15
6. 4 m; 10 m
7. Base = 12 cm;Altitude = 8 cm
8. 15 km, 20 km
9. 20 or 40
10. 9 kmph
EXERCISE - 5.3
1. (i)
1 33 1 33
,
4 4
, ∗ , ,
(ii)
3
2
,
,
3
2
,
(iii)
3
2,
5
,
(iv) ,1, ,5
2. (i)
1 33 1 33
,
4 4
, ∗ , ,
(ii)
3
2
,
,
3
2
,
(iii)
3
2,
5
,
(iv) ,1, ,5
3. (i)
3 13
2
,
,
3 13
2
∗
(ii) 1, 2
4. 7 years

Class-X Mathematics
378
5. Maths =12, English = 18 (or) Maths = 13, English= 17
6. 120 m; 90 m
7. 18, 12; ,18, ,12
8. 40 kmph
9. 15 hours, 25 hours
10. Speed ofthe passenger train = 33 kmph
Speed ofthe express train = 44 kmph
11. 18 m; 12 m
12. 4 seconds
13. 13 sides; No
EXERCISE - 5.4
1. (i) Realrootsdo not exist
(ii) Equalroots;
2
3
,
2
3
(iii) Distinct roots;
3 3
2
∗
,
3 3
2
,
2. (i) 2 6
k < ° (ii) 6
k <
3. Yes; 40 m; 20 m
4. No
5. Yes; 20 m; 20 m
EXERCISE - 6.1
1. (i) AP (ii) Not AP (iii) AP (iv) Not AP
2. (i) 10, 20, 30, 40 (ii) ,2, ,2, ,2, ,2
(iii) 4, 1, ,2, ,5 (iv) ,1,
1
2
, , 0,
1
2
(v) ,1.25, ,1.5, ,1.75, ,2
3. (i) a1
= 3; d = ,2 (ii) a1
= ,5; d = 4
(iii) 1
1
3
a < ;
4
3
d < (iv) a1
= 0.6; d = 1.1

Answers 379
4. (i) Not AP
(ii) AP, next three terms = 4,
9
2
, 5
(iii) AP, next three terms = ,9.2, ,11.2, ,13.2
(iv) AP, next three terms = 6, 10, 14
(v) AP, next three terms = 3 + 4 2 , 3 + 5 2 , 3 + 6 2
(vi) Not AP
(vii) AP, next three terms = ,16, ,20, ,24
(viii) AP, next three terms =
1
2
,
,
1
2
,
,
1
2
,
(ix) Not AP
(x) AP, next three term= 5a, 6a, 7a
(xi) Not AP
(xii) AP, next three terms = 50 , 72 , 98
(xiii) Not AP
EXERCISE - 6.2
1. (i) a8
= 28 (ii) d = 2 (iii) a = 46
(iv) n = 10 (v) an
= 3.5
2. (i) ,77 (ii) 22
3. (i) a2
= 14
(ii) a1
= 18; a3
= 8
(iii) a2
=
13
2
; a3
= 8
(iv) a2
= ,2; a3
= 0; a4
= 2; a5
= 4
(v) a1
= 53; a3
= 23; a4
= 8; a5
= ,7
4. 16th
term
5. (i) 34 (ii) 27

Class-X Mathematics
380
6. No 7. 178 8. 5 9. 1
10. 100 11. 128 12. 60 13. 13
14. AP = 4, 10, 16, .... 15. 158
16. ,13, ,8, ,3 17. 11
EXERCISE - 6.3
1. (i) 245 (ii) ,180 (iii) 5505 (iv)
33 13
1
20 20
<
2. (i)
2093 1
1046
2 2
< (ii) 286 (iii) ,8930
3. (i) n = 16 , 440 (ii) 13
7
, 273
3
d S
< <
(iii) a = 4, S12
= 246 (iv) d = –1, a10
= 8
(v) n = 5; a5
= 34 (vi) n = 7; a = ,8
(vii) a = 4
4. n = 38; S38
= 6973
5. 5610
6. n2
7. (i) 525 (ii) ,465
8. S1
= 3; S2
= 4; a2
= 1; a3
= ,1; a10
= ,15
an
= 5 ,2n
9. 4920 10. 160, 140, 120, 100, 80, 60, 40
11. 234 12. 143 13. 16 14. 370
EXERCISE - 6.4
1. (i) No (ii) No (iii) Yes
2. (i) 4, 12, 36, .... (ii)
5 5
5, , ,...
5 25
(iii) 81, ,27, 9, .... (iv)
1 1 1
, , , ......
64 32 16

Answers 381
3. (i) Yes; 32, 64, 128 (ii) Yes,
1 1 1
, ,
24 48 96
, ,
(iii) No (iv) Yes; -54, -162, -148 (v) No
(vi) Yes; ,81, 243, ,729 (vii) Yes; 2 3 4
1 1 1
, , , ......
x x x
(viii) Yes; 16, 32 2, 128
, , (ix) Yes; 0.0004, 0.00004, 0.000004
4. ,4
EXERCISE - 6.5
1. (i) r =
1
2
;
1
1
3
2
n
n
a
,
æ ö
÷
ç
< ÷
ç ÷
ç
è ø
(ii) r = ,3; an
= 2(,3)n-1
(iii) r = 3; an
= (,1)(3)n-1
(iv) r =
2
5
;
1
2
5
5
n
n
a
,
æ ö
÷
ç
< ÷
ç ÷
ç
è ø
2. a10
= 510
; an
= 5n
3. (i) 4
1
3
(ii) 4
4
3
,
4. (i) 5 (ii) 12 (iii) 7
5. 3 ´ 210
= 3072 6.
9 3
, , 1, ....
4 2
7. 5
EXERCISE - 7.1
1. (i) 2 2 (ii) 4 2 (iii) 5 2 (iv) 2 2
2 a b
∗
2. 39
3. Points arenot collinear 4. AB = BC = 37; AC = 2
5. AB = BC = CD = DA= 3 2 AC = BD = 6 (vertices of a square)
6. AB = BC = CA = 2a (vertices of equilateraltriangle)
7. AB = CD = 313 , BC =AD = 104 , AC ¹ BD (vertices of parallogram)
8. AB = BC = CD = DA= 90 , AC ¹ BD (vertices of a thombus), 72 Sq. units
9. (i) Square (ii) Trepezium (iii) Parallelogram
10. (,7, 0) 11. 7 or ,5

Class-X Mathematics
382
12. 3 or ,9 13. 4 5 units
14. AB = 5, BC = 10, AC = 15, AB + BC = AC = 15 (can not form the triangle)
15. x + 13y – 17 = 0
EXERCISE - 7.2
1. (1, 3) 2.
5
2,
3
æ ö
, ÷
ç ÷
ç ÷
ç
è ø
and
7
0,
3
æ ö
, ÷
ç ÷
ç ÷
ç
è ø
3. 2 : 7 4. x = 6 ; y = 3
5. (3, ,10) 6.
2 20
,
7 7
æ ö
, , ÷
ç ÷
ç ÷
ç
è ø
7.
3 9
3, , ( 2, 3), 1,
2 2
æ ö æ ö
÷ ÷
ç ç
, , ,
÷ ÷
ç ç
÷ ÷
ç ç
è ø è ø
8. ∋ (
13 7
1, , 1, , 0,5
2 2
æ ö æ ö
÷ ÷
ç ç,
÷ ÷
ç ç
÷ ÷
ç ç
è ø è ø
9.
5 5
,
5 5
a b a b
æ ö
, ∗ ÷
ç ÷
ç ÷
ç
è ø
10. (i)
2
, 2
3
æ ö
÷
ç ÷
ç ÷
ç
è ø
(ii)
10 5
,
3 3
æ ö
, ÷
ç ÷
ç ÷
ç
è ø
(iii)
2 5
,
3 3
æ ö
, ÷
ç ÷
ç ÷
ç
è ø
EXERCISE - 7.3
1. (i)
21
2
sq. units (ii) 32 sq. units (iii) 3 sq. units
2. (i) K = 4 (ii) K = 3 (iii) K =
7
3
3. 1 sq. unit ; 4 : 1 4. 28 sq. units 5. 6 sq. units
EXERCISE - 7.4
1. (i) 6 (ii) 3 (iii)
4b
a
(iv)
b
a
,
(v) -5 (vi) 0 (vii)
1
7
(viii) 1
,
EXERCISE - 8.2
1. (ii) DE = 2.8 cm
2. 8 cm

Answers 383
3. x = 5 cm and y =
13
2
16
cm or 2.8125 cm
4. 1.6 m 8. 16 m
EXERCISE - 8.3
1. 1: 4 2.
2 1
1
,
4. 96 cm2
6. 3.5 cm
EXERCISE - 8.4
8. 6 7 m 9. 13 m 12. 1: 2
EXERCISE - 9.1
1. (i) One (ii) Secant ofa circle (iii) Infinite
(iv) Point ofcontact (v) Infinite (vi) Two
2. PQ = 12 cm 4. 12 cm
EXERCISE - 9.2
1. (i) d (ii) a (iii) b (iv) a (v) c
2. 8 cm 4. AB = 15 cm, AC = 9 cm
5. 8 cm each 6. 2 5 cm 9. Two
EXERCISE - 9.3
1. (i) 28.5 cm2
(ii) 285.5 cm2
2. 88.368 cm2
3. 1254.96 cm2
4. 57 cm2
5. 10.5 cm2
6. 6.125 cm2
7. 102.67 cm2
8. 57 cm2
EXERCISE - 10.1
1. 5500 cm2
2. 184800 cm2
3. 264 c.c.
4. 1 : 2 5. 21
7. 21,175 cm3
8. 188.57 m2
9. 37 cm

Class-X Mathematics
384
EXERCISE - 10.2
1. 103.62 cm2
2. 1156.57 cm2
3. 298.57 mm2
4. 96 cm2
5. ` 827.20 6. 2 : 3 : 1
7.
2
6
4
x
æ ö
ο ÷
ç ∗ ÷
ç ÷
ç
è ø
sq. units 8. 374 cm2
EXERCISE - 10.3
1. 693 kg 2. Slant Height of cone(l) = 22.14 cm; Surface area of toy = 795.08 cm2
3. 89.83 cm3
4. 616 cm3
5. 309.57 cm3
6. 150 7. 523.9 cm3
EXERCISE - 10.4
1. 2.74 cm 2. 12 cm 3. 2.5 m
4. 5 m 5. 10 6. 400
7. 100 8. 672
EXERCISE - 11.1
1. sinA=
15
17
; cos A=
8
17
; tanA=
15
8
2.
527
168
3. cos π =
7
25
; tanπ =
24
7
4. sinA=
5
13
; tanA=
5
12
5. sinA=
4
5
; cos A=
3
5
7. (i)
49
64
(ii)
8 113
7
∗
8. (i) 1 (ii) 0
EXERCISE - 11.2
1. (i) 2 (ii)
3
4 2
(iii) 1
(iv) 2 (v) 1

Answers 385
2. (i) c (ii) d (iii) c
3. 1 4. Yes
5. QR = 6 3 cm; PR = 12 cm
6. Ð YXZ = 60º; Ð YXZ = 30º 7. Not true
EXERCISE - 11.3
1. (i) 1 (ii) 0 (iii) 0
(iv) 1 (v) 1
3. A = 36º 6. cos 15º + sin 25º
EXERCISE - 11.4
1. (i) 2 (ii) 2 (iii) 1
6. 1 8. 1 9.
1
p
EXERCISE - 12.1
1. 15 m 2. 6 3 m 3. 4 m
4. 30º 5. 34.64 m 6. 4 3 m
7. 4.1568 m 8. 300 3 m 9. 15 m 10. 7.5 cm2
EXERCISE - 12.2
1. Height of thetower = 5 3 m; Width of the road = 5 m
2. 32.908 m 3. 1.464 m 4. 19.124 m
5. 7.608 m 6. 10 m 7. 51.96 feets; 30 feets 8. 6 m
9. 200 m/sec. 10. 1:3
EXERCISE - 13.1
1. (i) 1 (ii) 0, Impossible event (iii) 1, Sure event
(iv) 1 (v) 0, 1
2. (i) Yes (ii) Yes (iii) Yes (iv) Yes
3. 0.95 4. (i) 0 (ii) 1

Class-X Mathematics
386
5. 1 1
, , 1, 0
13 3
6. 0.008 7. (i)
1
2
(ii)
1
2
(iii)
1
2
8.
1
26
EXERCISE - 13.2
1. (i)
3
8
(ii)
5
8
2. (i)
5
17
(ii)
4
17
(iii)
13
17
3. (i)
5
9
(ii)
17
18
4.
5
13
5. (i)
1
8
(ii)
1
2
(iii)
3
4
(iv) 1
6. (i)
1
26
(ii)
3
13
(iii)
1
26
(iv)
1
52
(v)
1
4
(vi)
1
52
7. (i)
1
5
(ii)(a)
1
4
(b) 0
8.
11
12
9. (i)
1
5
(ii)
15
19
10. (i)
9
10
(ii) (a)
1
10
(b)
1
5
11.
11
84
12. (i)
31
36
(ii)
5
36
13. (i)
1 2 3 4 5 6 5 4 3 2 1
, , , , , , , , , ,
36 36 36 36 36 36 36 36 36 36 36
(ii) No
14.
3
4
15. (i)
25
36
(ii)
11
36

Answers 387
16.
Sum on2 dice 2 3 4 5 6 7 8 9 10 11 12
Probability
1
36
1
18
1
12
1
9
5
36
1
6
5
36
1
9
1
12
1
18
1
36
17. (i)
1
2
(ii)
1
2
EXERCISE - 14.1
1. 8.1 plants. We have used direct method because numericalvalues ofxi
and fi
are small.
2. ` 313 3. f = 20 4. 75.9
5. 22.31 6. ` 211 7. 0.099 ppm
8. 49 days 9. 69.43%
EXERCISE - 14.2
1. Mode = 36.8 years, Mean = 35.37 years, Maximumnumber ofpatientsadmitted in the
hospital are ofthe age 36.8 years (approx.), while on an average the age of a patient
admitted to the hospitalis 35.37 years.
2. 65.625 hours
3. Modal monthlyexpenditure = ` 1847.83, Mean monthlyexpenditure = ` 2662.5.
4. Mode : 30.6, Mean = 29.2. Most states/U.T. have a student teacher ratio of30.6 and
on an average, this ratio is 29.2.
5. Mode = 4608.7 runs.
6. Mode = 44.7 cars
EXERCISE - 14.3
1. Median = 137 units, Mean = 137.05 units, Mode = 135.76 units.
The three measures are approximatelythe same in thiscase.
2. x = 8, y = 7

Class-X Mathematics
388
3. Median age = 35.76 years
4. Median length= 146.75 mm
5. Median life = 3406.98 hours
6. Median = 8.05, Mean = 8.32, Modal size = 7.88
7. Median weight = 56.67 kg
EXERCISE - 14.4
1. Dailyincome(in`) Cumulativefrequency
Less than 300 12
Less than 350 26
Less than 400 34
Less than 450 40
Less than 500 50
2. Draw the ogive by plotting the points : (38, 0), (40, 3), (42, 5), (44, 9), (46, 14),
(48, 28), (50, 32) and (52, 35). Here
2
n
=17.5. Locate the point on the ogive whose
ordinate is 17.5. The x-coordinate ofthis point willbe the median.
3. Productionyield (kg/ha) Cumulativefrequency
Now, draw the ogive by plotting the points : (50, 100), (55, 98), (60, 90), (65, 78),
(70, 54) and (75, 16).
Draw ogive byplotting the points :
(300, 12), (350, 26), (400, 34),
(450, 40) and (500, 50)

389
Dear teachers,
The State Government of has decided to revise the curriculum of all the subjects based on
State Curriculum Framework (SCF-2011). The framework emphasises that all children must learn
and the mathematics learnt at school must be linked to the life and experience of them. The NCF-
2005, the position paper on Mahtematics of the NCERT and the Govt. of emphasise, building
understanding and developing the capability, exploration and inclination to mathematize experiences.
This would become more possible at the secondary level. We have consolidated the basic framework
of mathematics in class-IX and now we are at level of completion of secondary level of mathematics.
In previousclasses, we have encouraged the studentsfor greater abstraction and formal mathematical
formulation. We made them to deal with proofs and use mathematical language. It is important to
recognise that as we go forward the language- in which mathematical arguments and statements-
are presented would become even more symbolic and terse. It is therefore important in this class to
help children become comfortable and competent in using mathematical ideas. In class-X, we will
make all such idea at level of total abstraction.
It would be important to consider all the syllabi from class-VI to X while looking at teaching
class X. Thenature and extent of abstraction and use of mathematical language is gradually increasing.
The program here would also become axiomatic and children must be slowly empowered to deal with
that. One of the major difficultieschildren have in moving forward and learning secondary mathematics
is their inability to deal with the axiomatic nature and language of symbols. They need to have an
apportunity to learn and develop these perspectives by engaging with, as a team. Peer support in
overcoming the difficulties is critical and it would be important to put them in a group to think, discuss
and solve problems. When children will learn such things in class-X, it will be helpful for them in future
mathematical learning also.
The syllabus is based on the structural approach, laying emphasis on the discovery and
understanding of basic mathematical concepts and generalisations. The approach is to encourage
participation and discussion in classroom activities.
The syllabus in textbook of Class-X Mathematics has been divided broadly into six areas
Number System, Arithmetic, Algebra, Geometry, Trigonometry, Statistics and Coordinate Geometry.
Teaching of the topics related to these areas will develop the skills such as problem solving, logical
thinking, mathematical communication, representing data in various forms, using mathematics as
one of the disciplines of study and also in daily life situations.
This text book attempts to enhance this endeavor by giving higher priority and space to
opportunities for contemplations. There is a scope for discussion in small groups and activities
required for hands on experience in the form of ‘Do this’and ‘Try this’. Teacher’s support is needed in
setting the situations in the classroom and also for development of interest in new book.
Exercises in ‘Do This’ and ‘Try This’ are given extensively after completion of each concept.
The problems which are given under ‘Do This’ are based on the concept taught and ‘Try This’ problems
Note to the Teachers

390
are intended to test the skills of generalization of facts, ensuring correctness and questioning. ‘Think,
Discuss and Write’ has given to understand the new concept between students in their own words.
Entire syllabus in class-X Mathematics is divided into 14 chapters with an appendix, so that
a child can go through the content well in bit wise to consolidate the logic and enjoy the learning of
mathematics. Colourful pictures, diagrams, readable font size will certainly help the children to adopt
the contents and care this book as theirs.
Chapter-1 : Real number, we are discussing about the exploration of real numbers in which
the brief account of fundamental theorem of arithmetic, rational numbers their decimal expansion
and non-terminating recurring rational numbers has given. Here we are giving some more about the
irrational numbers. In this chapter, first time we are introducing logarithms in this we are discussing
about basic laws of logarithms and their application.
Chapter-2 : Sets, this is entirely a new chapter at the level of secondary students. In old
syllabus it was there but here we are introducing it in class X. This chapter is introduced with wide
variety of examples which are dealing about the definition of sets, types of set, Venn diagrams,
operations of sets, differencesbetween sets. In thischapter we dealt about howto develop a common
understanding of sets. How can you make set of any objects?
Chapter-3 : Polynomials, we are discussing about the fact "what are polynomials?" and
degree and value of polynomials come under it. This time we look at the graphical representation of
linear equations and quadratic equations. Here we are taking care of zeros and coefficients of a
polynomial & their relationship. We also start with cubic polynomials and division algorithm of
polynomials.
Chapter-4 : Pair of linear equations in two variables, we start the scenario with discussing
about finding of unknown quantities and use of two equations together. Solution of pair of linear
equations in two variables with the help of graphical and algebraic methods has done. Here we have
illustrated so many examples to understand the relation between coefficients and nature of system
of equations. Reduction of equation to two variable linear equation has done here.
The problem is framed in such a way to emphasis the correlation between various chapters
within the mathematics and other subjects of daily life of human being. This chapter links the ability
of finding unknown with every day experience.
Chapter-5 : Quadratic equations, states the meaning of quadratic equation and solution of
quadratic equation with the factorizations completion of squares. Nature of roots is defined here with
the use of parabola.
Chapter-6 : Progressions, we have introduced this chapter first time on secondary level. In
this chapter use are taking about arithmetic progressions and geometric progressions. How the
terms progressing arithmetically and geometrically in progressions discussed. The number of terms,
nth terms, sum of terms are stated in this chapter.
Chapter-7 : Coordinate geometry, deals with finding the distance between two points on
cartesion plane, section formula, centroid of a triangle and tsisectional points of a line. In this, we are
also talking about area of the triangle on plane and finding it with the use of 'Heron's formula'. The
slope on straight line is also introduced here.
Class-X Mathematics

391
We are keeping three chapters (8, 9, 10) in X mathematics book and all of them are having
emphasis on learning geometry using reasoning, intuitive understanding and insightful personal
experience of meanings. If helps in communicating and solving problemsand obtaining newrelations
among various plane figures. In chapter 9 Tangents and Secants to a circle, we have introduced the
new terms caked tangent and secant with their properties. We also discussed about the segment
and area of that which is formed by secant. Menstruations are presented in combination of solids
and finding of their volume and area.
We are keeping two newchapters(11 & 12) at second level for the first time. The applications
of trianglesare used with giving relation with the hypotenuse, perpendicular and base. These chapters
are the introduction of trigonometry which have very big role in high level studies and also in
determination of so many measurements. Applications of trigonometry are also given with brief idea
of using triangle.
Chapter-13 : Probability, is little higher level chapter than the last chapter which we have
introduced in class IX. Here we are taking about different terms of probability by using some daily life
situations.
Chapter-14 : Statistics, deals with importance of statistics, collection of statistical grouped
data, illustrative examples for finding mean, median and mode of given grouped data with different
methods. The ogives are also illustrated here again. In appendix, mathematical modeling is given
there with an idea about the models and their modeling methods.
The success of any course depends not only on the syllabus but also on the teachers and
the teaching methods they employ. It is hoped that all teachers are concerned with the improving of
mathematics education and they will extend their full co operation in this endeavour.
The production of good text books does not ensure the quality of education, unless the
teachers transact the curriculum in a way as it is discussed in the text book. The involvement and
participation of learner in doing the activities and problems with an understanding is ensured.
Students should be made to digest the concepts given in “What we have discussed”
completely. Teachers may prepare their own problems related to the concepts besides solving the
problems given in the exercises. So therefore it is hopped that the teachers will bring a paradigm shift
in the classroom process from mere solving the problems in the exercises routinely to the conceptual
understanding, solving of problems with ingenity.
Teaching learning strategies and the expected learning outcomes, have been developed
class wise and subject-wise based on the syllabus and compiled in the form of a Hand book to guide
the teachers and were supplied to all the schools. With the help of this Hand book the teachers are
expected to conduct effective teaching learning processes and ensure that all the students attain the
expected learning outcomes.
“Good luck for happy teaching”
Note to the Teachers

392
I. NUMBER SYSTEM (23 PERIODS)
(i) Real numbers (15 periods)
• More about rational and irrational numbers.
• Euclid Division lemma
• Fundamental theorem of Arithmetic - Statements.
• Proofsof results- irrationality of 2, 3 etc. and Decimal expansionsof rational numbers
in terms of terminating / non-terminating recurring decimals and vice versa.
• Properties of real numbers (after reviewing lookdone earlier and after illustrating and
motivating through examples)
• Introduction of logarithms
• Conversion of a number in exponential form to a logarithmic form
• Properties of logarithms loga a = 1; loga1 = 0
• Laws of logarithms
log xy = log x + log y; log
x
y
= log x - log y; logxn = n log x, log N
N
a
a =
• Standard base of logarithm and use of logarithms in daily life situations (not meant for
examination)
(ii) Sets (8 periods)
• Sets and their representations
• Empty set, Finite and infinite sets, universal set
• Equal sets, subsets, subsets of set of real numbers (especially intervals and notations)
• Venn diagrams and cardinality of sets
• Basic set operations - union and intersection of sets
• Disjoint sets, difference of sets
II. ALGEBRA (46 PERIODS)
(i) Polynomials (8 periods)
• Zeroes of a polynomial
• Geometrical meaning of zeroes of linear, quadratic and cubic polynomials using graphs.
• Relationship between zeroes and coefficients of a polynomial.
• Simple problems on division algorithm for polynomials with integral coefficients
(ii) Pair of Linear Equations in Two Variables (15 periods)
• Forming a linear equation in two variables through illustrated examples.
• Graphical representation of a pair of linear equations of different possibilities of solutions
/ in consistency.
• Algebraic conditions for number of solutions
• Solution of pair of linear equations in two variables algebraically - by Substitution, by
elimination.
• Simple and daily life problems on equations reducible to linear equations.
(iii) Quadratic Equations (12 periods)
• Standard form of a quadratic equation ax2 + bx + c = 0, a ¹ 0.
• Solutions of quadratic equations (only real roots) by factorisation and by completing the
square i.e. by using quadratic formula.
Syllabus

393
• Relationship between discribmenant and nature of roots.
• Problems related to day-to-day activities.
(iv) Progressions (11 periods)
• Definition of Arithmetic progression (A.P)
• Finding nth term and sum of first nterms of A.P.
• Geometric progression (G.P.)
• Find nth term of G.P.
III. GEOMETRY (33 PERIODS)
(i) Similar Triangles (18 periods)
• Similarly figures difference between congruency and similarity.
• Properties of similar triangles.
• (Prove) If a line is drawn parallel to one side of a triangle to intersect the other two sides
in distinct points, the other twosides are divided in the same ratio.
• (Motivate) If a line divides two sides of a triangle in the same ratio, then the line is
parallel to the third side.
• (Motivate) If in two triangles, the corresponding angles are equal, their corresponding
sides are proportional and the triangles are similar (AAA).
• (Motivate) If the corresponding sidesof two trianglesare proportional, their corresponding
angles are equal and the two triangles are similar (SSS).
• (Motivate) If one angle of a triangle isequal to one angle of another triangle and the sides
including these angles are proportional, the two triangles are similar (SAS).
• (Prove) The ratio of the areas of two similar triangles is equal to the ratio of the squares
on their corresponding sides.
• (Motivate) If a perpendicular isdrawn from the vertex of the right angle to the hypotenuse,
the triangles on each side of the perpendicular are similar to the whole triangle and to
each other.
• (Prove) In a right triangle, the square on the hypothenuse is equal to the sum of the
squares on the other two sides.
• (Prove) In a triangle, if the square on one side is equal to sum of the squares on the other
sides, the angles opposite to the first side is a right triangle.
• (Construction) Division of a line segment using Basic proportionality Theorem.
• (Construction)Atriangle similar to a given triangle as per given scale factor.
(ii) Tangents and secants to a circle (15 periods)
• Difference between tangent and secant to a circle
• Tangents to a circle motivated by chords drawn from points coming closer and closer to
the point
• (Prove) The tangent at any point on a circle is perpendicular to the radius through the
point contact.
• (Prove) The lengths of tangents drawn from an external point to a circle are equal.
• (Construction) A tangent to a circle through a point given on it.
• (Construction) Pair of tangents to a circle from an external point.
• Segment of a circle made by the secant.
• Finding the area of minor/major segment of a circle.
Syllabus

394
IV. COORDINATE GEOMETRY
• Review the concepts of coordinate geometry by the graphs of linear equations.
• Distance between two points P(x1, y1) and Q(x2, y2)
2 2
2 1 2 1
PQ ( ) ( )
x x y y
= - - -
• Section formula (internal division of a line segment in the ratio m : n).
• Area of triangle on coordinate plane.
• Slope of a line joining two points
V. TRIGONOMETRY (23 PERIODS)
(i) Trigonometry (15 periods)
• Trigonometric ratiosof an actute angle of a right angled triangle i.e. sine, cosine, tangent,
cosecant, secant and cotangent.
• Values of trigonometric ratios of 30o, 45o and 60o (with proofs).
• Relationship between the ratios and trigonometric ratios for complementary angles.
• Trigonometric Identities.
(i) sin2 A + cos2 A = 1, (ii) 1 + tan2 A = sec2 A, (iii) cot2 A + 1 = cosec2 A
(ii) Applications of Trigonometry (8 periods)
• Angle of elevation, angle of depression.
• Simple and daily life problems on heights and distomces
• Problemsinvolving not more than two right triangles and angles of elevation / depression
confined to 30o, 45o and 60o.
VI. MENSURATION (10 PERIODS)
(i) Surface Areas and Volumes
• Problems on finding surface area and volumes of combinations of any two of the following
i.e. cubes, cubiods, right cicular cylinders, cones spheres and hemispheres.
• Problemsinvolving converting one type of mettalic solid into anothers and finding volumes
and other mixed problems involving not more than two different solids.
VII. DATA HANDLING (25 PERIODS)
(i) Statistics
• Revision of mean, median and mode of ungrouped (frequency distribution) data.
• Understanding the concept of Arithmetic mean, median and mode for grouped (classified)
data.
• Simple problems on finding mean, median and mode for grouped/ungrouped data with
different methods.
• Usage and different values of central tendencies through ogives.
(ii) Probability (10 periods)
• Revision of concept and definition of probability.
• Simple problems (day to day life situation) on single events using set notation.
• Concept of complimentary events.
APPENDIX
Mathematical Modeling (8 periods)
• Concept of Mathematical modelling
• Discussion of broad stages of modelling-real life situations (Simple interest, Fair
installments payments etc. ....)
Class-X Mathematics

395
Academic Standards : Academic standards are clear statements about what students
must know and be able to do. The following are categories on the basis of which we lay
down academic standards.
Areas of Mathematics Content
1. Problem Solving Using concepts and procedures to solve mathematical
problems like following:
a. Kinds of problems :
Problems can take various forms- puzzles, word
problems, pictorial problems, procedural problems,
reading data, tables, graphs etc.
• Reads problems.
• Identifies all pieces of information/data.
• Separates relevant pieces of information.
• Understanding what concept is involved.
• Recalling of (synthesis of) concerned procedures,
formulae etc.
• Selection of procedure.
• Solving the problem.
• Verification of answers of raiders, problem based
theorems.
b. Complexity :
The complexity of a problem is dependent on-
• Making connections( as defined in the connections
section).
• Number of steps.
• Number of operations.
• Context unraveling.
• Nature of procedures.
2. Reasoning Proof • Reasoning between various steps (involved invariably
conjuncture).
• Understanding and making mathematical
generalizations and conjectures
Academic Standards - High School

396
• Understands and justifies procedures
• Examining logical arguments.
• Understanding the notion of proof
• Uses inductive and deductive logic
• Testing mathematical conjectures
3. Communication • Writing and reading, expressing mathematical
notations (verbal and symbolic forms)
Example : 3+4=7
n1
+n2
= n2
+n1
Sum of angles in triangle = 180°
• Creating mathematical expressions
4. Connections • Connecting concepts within a mathematical domain-
for example relating adding to multiplication, parts of
a whole to a ratio, to division. Patterns and symmetry,
measurements and space.
• Making connections with daily life.
• Connecting mathematics to different subjects.
• Connecting concepts of different mathematical
domains like data handling and arithmetic or arithmetic
and space.
• Connecting concepts to multiple procedures.
5. Visualization & • Interprets and reads data in a table, number line,
Representation pictograph, bar graph, 2-D figures, 3- D figures,
pictures.
• Making tables, number line, pictograph, bar graph,
pictures.
• Mathematical symbols and figures.
Class-X Mathematics

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