Class 10 RD Sharma Mathematics Solutions - Chapter 4 Triangles - Exercise 4.6 | Set 2

Chapter 4 of RD Sharma's Class 10 mathematics book focuses on "Triangles" which is a fundamental concept in the geometry. This chapter deals with the various properties and theorems related to triangles such as the congruence similarity and Pythagorean theorem. Understanding these concepts is essential for the solving problems related to the triangles and other advanced topics in geometry.

Important Theorems and Formulas: Triangles

A triangle is a polygon with the three edges and three vertices. The sum of the interior angles of the triangle is always 180 degrees. The Triangles can be classified based on their sides and angles. Understanding the properties and types of the triangles is crucial for the solving various geometric problems.

Pythagoras Theorem

In a right-angled triangle the square of the hypotenuse is equal to sum of the squares of the other two sides.

c2 = a2 + b2

where c is the hypotenuse and a and b are the other two sides.

Area of a Triangle

The area of a triangle is given by:

Area = \frac{1}{2} \times \text{base} \times \text{height}

This formula is used to find the area of the triangle when the base and height are known.

Similarity of Triangles

Two triangles are similar if:

• Corresponding angles are equal.
• Corresponding sides are in proportion.

Basic Proportionality Theorem (Thales' Theorem)

If a line is drawn parallel to the one side of a triangle to the intersect the other two sides in distinct points the other two sides are divided in the same ratio.

Heron’s Formula

To find the area of a triangle when all three sides are known:

Area = \sqrt{s(s-a)(s-b)(s-c)}

where s = \frac{a+b+c}{2} is the semi-perimeter of the triangle.

Class 10 RD Sharma Mathematics Solutions - Exercise 4.6 | Set 2

Question 11. The areas of two similar triangles are 121 cm² and 64 cm² respectively. If the median of the first triangle is 12.1 cm, find the corresponding median of the other.

Solution:

Let us consider ∆ABC and ∆DEF, AL and DM are the medians of ∆ABC and ∆DEF 

It is given that the area of ∆ABC = 121 cm² and area of ∆DEF = 64 cm²

AL = 12.1 cm

Let us assume DM = x cm

Given that, ∆ABC ~ ∆DEF

So, 

ar(∆ABC)/ar(∆DEF) = AL²/DM² 

= 121/64 = (12.1)²/x²

11/8 = 12.1/x 

⇒ x = (8 × 12.1)/11 = 8.8

Hence, the median of the second triangle is 8.8cm 

Question 12. In ∆ABC ~ ∆DEF such that AB = 5 cm and (∆ABC) = 20 cm² and area (∆DEF) = 45 cm², determine DE.

Solution:

Given that,  

area (∆ABC) = 20 cm²

area (∆DEF) = 45 cm²

AB = 5 cm

Let us consider DE = x cm

Also, given that ∆ABC ~ ∆DEF

ar(∆ABC)/ar(∆DEF) = AB²/DE² 

⇒20/45 = (5)²/x²

⇒20/45 = 25/x² 

⇒x² = (25 × 45)/20 = 225/4 = (15/2)²

x = 15/2 = 7.5 

DE = 7.5cm

Question 13. In ∆ABC, PQ is a line segment intersecting AB at P and AC at Q such that PQ || BC and PQ divide ∆ABC into two parts equal in area. Find BP/AB.

Solution:

It is given that, in ∆ABC, PQ || BC and line PQ divide the ∆ABC into two parts 

∆APQ and trap. BPQC equally

i.e., area ∆APQ = area BPQC

Now we have to find BP/AB.

As we know that PQ||BC

So, ∆APQ ∼ ∆ABC

⇒ ar.(∆APQ)/ar.(∆ABC) = AP²/AB²

⇒ ar.(∆ABC)/ar.(∆APQ) = AB²/AP²

2/1 = AB²/AP²

{area ∆APQ = area trap. BPQC

 Area ∆ABC = 2area (∆APQ)}

⇒ AB/AP = √2/1 

⇒√2 AP = AB = AP + PB

⇒√2AP - AP = PB

⇒(√2 - 1)AP = PB

BP/AP = (√2 - 1)/1

Question 14. The areas of two similar triangles ABC and PQR are in the ratio 9 : 16. If BC = 4.5 cm, find the length of QR. 

Solution:

Given that, area (∆ABC) : area (∆PQR) = 9 : 16

∆ABC ~ ∆PQR

and BC = 4.5 cm

Let us considered QR = x cm

As we know that ∆ABC ~ ∆PQR

ar.(∆ABC)/ar.(∆PQR) = BC²/QR² ⇒ 9/16 = (4.5)²/x²

⇒ (3/4)² = (4.5/x)² ⇒ 4.5/x = 3/4

x = (4.5 × 4)/3 = 60

Hence, the length of QR is 6cm

Question 15. ABC is a triangle and PQ is a straight line meeting AB in P and AC in Q. If AP = 1 cm, PB = 3 cm, AQ = 1.5 cm, QC = 4.5 cm, prove that area of ∆APQ is one sixteenth of the area of ∆ABC.

Solution:

Given that, in ∆ABC, P and Q are two points on line AB and AC 

AP = 1 cm, PB = 3 cm, AQ = 1.5 cm and QC = 4.5 cm

Now, AP/PB = 1/3 and AQ/QC = 1.5/4.5 = 1/3

In ∆APQ and ∆ABC

AP/PB = AQ/QC

PQ||BC

Hence, ∆APQ ∼ ∆ABC

So, ar.(∆APQ)/ar.(∆ABC) = AP²/PB² = AP²/(AP + PB)²

 ar.(∆APQ)/ar.(∆ABC) = 1²/(1 + 3)² = 1/16

Hence, area of ∆APQ = 1/16 of area of ∆ABC

Question 16. If D is a point on the side AB of ∆ABC such that AD : DB = 3 : 2 and E is a point on BC such that DE || AC. Find the ratio of areas of ∆ABC and ∆BDE. 

Solution:

Given that in ∆ABC, D is a point on AB such that AD : DB = 3 : 2

DE||AC

In ∆BDE and ∆ABC

∠BDE = ∠A

∠DBE = ∠ABC

So, by AA, ∆BED ∼ ∆ABC

Therefore, ar.(∆ABC)/ar.(∆BDE) = AB²/BD² = (BD + AD)²/BD²

= (2 + 3)²/2² = 5²/2² = 25/4

Hence, the ratio of areas of ∆ABC and ∆BDE is 25:4

Question 17. If ∆ABC and ∆BDE are equilateral triangles, where D is the mid point of BC, find the ratio of areas of ∆ABC and ∆BDE.

Solution:

Given that ∆ABC and ∆DBE are equilateral triangles, where D is mid point of BC

So, BD = 1/2BC

Now area of ∆ABC

√3/4(side)² = √3/4BC²

and area of ∆DBE

√3/4(side)² = √3/4BD²

√3/4(side)² = √3/4(1/2BC)²

√3/4(side)² = √3/16(BC)²

So, the ratio between areas is

= area(∆ABC)/area(∆DBE) =\frac{\frac{\sqrt{3}}{4}BC^2}{\frac{\sqrt{3}}{16}BC^2}\\ =\frac{\sqrt{3}}{4}×\frac{16}{\sqrt{3}}=\frac{4}{1}

Hence, the ratio of areas of ∆ABC and ∆BDE is 4:1

Question 18. Two isosceles triangles have equal vertical angles and their areas are in the ratio 36 : 25. Find the ratio of their corresponding heights.

Solution:

Let us consider two triangles, ∆ABC and ∆XYZ and these triangles have equal vertical angle, i.e., ∠A and ∠X

And AD and XO is the heights of these triangles.

So, ∆ABC/∆XYZ = AB/AC = XY/XZ

In ∆ABC and ∆XYZ 

∠A = ∠X

AB/AC = XY/XZ

So, by SAS

∆ABC ~ ∆XYZ

So, ar(∆ABC)/ar(∆XYZ) = AD²/XO²

As we know that ar(∆ABC)/ar(∆XYZ) = 36/25

So, 

36/25 = AD²/XO²

6/5 = AD/XO

Hence, the ratio of their corresponding heights is 6:5

Question 19. In the figure, ∆ABC and ∆DBC are on the same base BC. If AD and BC intersect at O, prove that \frac{ar.(∆ABC)}{ar.(∆DBC)}=\frac{AO}{DO}

Solution:

Given that two ∆ABC and ∆DBC are on the same base BC as shown in the given figure

AC and BD intersect each other at O

Now, Draw AL ⊥ BC and DM ⊥BC

Prove: \frac{ar.(∆ABC)}{ar.(∆DBC)}=\frac{AO}{DO}

Proof: 

In ∆ALO and ∆DMO,

∠L =∠M = 90°

∠AOL = ∠DOM [Vertically opposite angles]

So, by AA, ∆ALO ∼ ∆DMO       

So, AL/DM = AO/DO 

Now \frac{ar(∆ABC)}{ar(∆DBC)}=\frac{\frac{1}{2}BC×AL}{\frac{1}{2}BC×DM}=\frac{AL}{DM}

But AL/DM = AO/DO      (Proved above)

So, \frac{ar.(∆ABC)}{ar.(∆DBC)}=\frac{AO}{DO} 

Hence proved

Question 20. ABCD is a trapezium in which AB || CD. The diagonals AC and BD intersect at O. Prove that

(i) ∆AOB ~ ∆COD

(ii) If OA = 6 cm, OC = 8 cm, find

(a)ar(∆AOB)/ar(∆COD)        (b) ar(∆AOD)/ar(∆COD)

Solution:

Given that ABCD is a trapezium in which AB || CD and the diagonals AC and BD intersect at O

Now, in the figure from point D, draw DL⊥AC

(i) In ∆AOB and ∆COD

∠AOB =∠COD            [Vertically opposite angles]

∠OAB =∠OCD           [Alternate angles]

So, by AA criterion

∆AOB ~ ∆COD         

(ii) Given that OA = 6 cm, OC = 8 cm

As we know that ∆AOB ~ ∆COD     

So, OA/OC = OB/OD = AB/CD

(a) ar(∆AOB)/ar(∆COD) = AO²/OC²

= 6²/8² = 36/64 = 9/16 

Therefore, ar(∆AOB)/ar(∆COD) = 9/16 

(b) As we know that ∆AOD and ∆COD have their bases on the same line and their vertex A is common

Therefore,  ar(∆AOD)/ar(∆COD) = AO/OC = 6/8 = 3/4   

Question 21. In ∆ABC, P divides the side AB such that AP : PB = 1 : 2. Q is a point in AC such that PQ || BC. Find the ratio of the areas of ∆APQ and trapezium BPQC.

Solution:

Given that, ABC is a triangle, in which P divides the side AB such that 

AP : PB = 1 : 2. Q is a point in AC such that PQ || BC

In ∆APQ and ∆ABC

∠APQ = ∠B

∠PAQ = ∠BAC

So, by AA criterion

∆APQ ∼ ∆ABC

So, 

ar(∆APQ)/ar(∆ABC) = (AP)²/(AB)²

ar(∆APQ)/ar(∆ABC) = (1)²/(1 + 2)² = (1)²/(3)² = 1/9

9 ar(∆APQ) = ar(∆ABC)

9 ar(∆APQ) = ar(∆APQ) + ar(trap. BPQC)

9 ar(∆APQ) = ar(trap BPQC)

ar(∆APQ)/ar(trap BPQC) = 1/9

Hence, the ratio of the areas of ∆APQ and trapezium BPQC is 1:9

Question 22. AD is an altitude of an equilateral triangle ABC. On AD as base, another equilateral triangle ADE is constructed. Prove that Area (∆ADE) : Area (∆ABC) = 3 : 4. 

Solution:

Given that AD is an altitude of an equilateral triangle ABC. 

On AD as base, another equilateral triangle ADE is constructed

Prove: Area (∆ADE) : Area (∆ABC) = 3 : 4

Proof:

Area of ∆ABC = √3/4 BC²

and AD = √3/2 BC 

Area of ∆ADE = √3/4 AD²

= √3/4 (√3/2 BC)² = 3√3/16 BC²

So, the ratio of area (∆ADE):area(∆ABC) = 3√3/16 BC² : √3/4 BC²

= 3/4:1 = 3:4

Read more,

• Types of Triangles, Classifications, Practice Questions and FAQs
• Triangles in Geometry
• Class 10 RD Sharma Mathematics Solutions - Chapter 4 Triangles - Exercise 4.5
• Class 10 RD Sharma Mathematics Solutions - Chapter 4 Triangles - Exercise 4.6 | Set 1
• Class 10 RD Sharma Mathematics Solutions - Chapter 4 Triangles - Exercise 4.7

Conclusion

In this exercise, we explored various properties and theorems related to the triangles such as the Pythagoras theorem the concept of the similar triangles and use of the Basic Proportionality Theorem. Understanding these foundational principles is crucial as they form the basis for the solving more complex problems in geometry. The Regular practice and application of these formulas will help in the mastering the concepts of triangles.