Module 3 similarity

Module 3
Similarity
What this module is about
This module is about similarities on right triangles. As you go over the
exercises you will develop skills in applying similarity on right triangles and solve
for the missing lengths of sides using the famous Pythagorean theorem.
What you are expected to learn
1. Apply AA similarity on Right triangles
2. In a right triangle, the altitude to the hypotenuse separates the triangle
into two triangles each similar to the given triangle and similar to each
other.
3. On a right triangle, the altitude to the hypotenuse is the geometric
mean of the segments in which it divides, each leg is the geometric
mean of the hypotenuse and the segment of the hypotenuse adjacent
to it.
4. Pythagorean Theorem and its application to special right triangles
How much do you know
Use the figure to answer each of the following:
1. What is the hypotenuse of rt. ∆ABC?
C
BA
D

2
2. If ∠ C is the right angle of ∆ABC and CD ⊥ AB then ∆ABC ∼ ∆BDC ∼ ___.
3. Complete the proportion:
BDCD
AD ?
=
4. In rt. ∆PRO, ∠ R is a right angle OR = 24 and PO = 26,
find PR:
5. In a 30°-60°-90° triangle the length of the hypotenuse is 14. Find the
length of the longer leg.
6. In a 30°-60°-90° triangle, the length of the hypotenuse is 11
2
1
. Find the
length of the shorter leg.
7. In a 45° - 45° - 90° triangle, the length of the hypotenuse is 16. Find the
length of a leg.
8. Find the length of the altitude of an equilateral triangle if the length of a
side is 6.
9. Find the length of the diagonal of a square if the length o a side is 10 cm.
10. ∆BAC is a right triangle ∠C is right angle CD ⊥ AB. Find CD if AD = 14,
DB = 6
O
P R
24
26
C
B
D
A
14 6

3
What will you do
Lesson 1
Similarity on Right Triangle
Let us recall the AA Similarity Theorem.
Given a correspondence between the vertices of two triangles. If two pairs
of corresponding angles are congruent, then the triangles are similar.
From the theorem, if ABC ↔ RST and ∠A = ∠R, ∠B = ∠S then ∆ABC ∼
∆RST. We can apply this theorem to prove another theorem, this time in a right
triangle.
Theorem: In a right triangle, the altitude to the hypotenuse separates the
triangle into to two triangles each similar to the given triangle and
similar to each other.
Given: Right ∆ABC with altitude CP
Prove: ∆ACP ∼ ∆CBP ∼ ∆ABC
To prove this theorem, we apply the AA Similarity Theorem
A
P
B
C
S
TR
B
A C

4
Examples:
If you are given ∆PRT a right triangle and RM an altitude to the hypotenuse
then we can have three pairs of similar triangles.
∆RMP ∼ ∆TRP
∆TMR ∼ ∆TRP
∆RMP ∼ ∆TMR
Try this out
A. Use the figure to answer each of the following:
1. Name the right triangle of ∆ABC
2. What is the altitude to the hypotenuse of ∆ABC?
3. Name the hypotenuse of ∆ABC
4. Two segments of the hypotenuse
Are AD and ____.
5. The hypotenuse of ∆BCD if
CD ⊥ AB is ____.
6. Name the right angle of ∆ACD
7. Name the hypotenuse of right ∆BCD
8. ∆ADC ∼ _____
9. ∆ABC ∼ _____
10.∆ABC ∼ _____
B. Name the pairs of right triangles that are similar.
1.
2.
3.
R
P
M
T
D
C
BA
S
R T
O

5
4.
5.
6.
C. Use the figure at the right.
1. Name all the right triangles.
2. In ∆ABC, name the altitude to the hypotenuse.
3. Name the hypotenuse in ∆ADC.
4. Name the hypotenuse of ∆ACB.
One of the segments shown is an altitude to the hypotenuse of a right
triangle. Name the segment.
5. 6. 7.
Name the three pairs of similar triangles:
8.
9.
10.
A
D
C B
B
CA
D
M
N
OP
M
S
RO
I D
J
KG F
H
E

6
Lesson 2
Geometric Mean in Similar Right Triangles
The previous theorem states that:
In a right triangle, the altitude to the hypotenuse divides the triangle into
similar triangles, each similar to the given triangle.
If : ∆ACB is right with ∠C, the right angle
CD is the altitude to the hypotenuse AB.
Then : ∆ADC ∼ ∆ACB
∆CDB ∼ ∆ACB
∆ADC ∼ ∆CDB
Corollary: 1. In a right triangle, the altitude to the hypotenuse is the geometric
mean of the segments into which divides the hypotenuse
In the figure:
DB
CD
CD
AD
=
Corollary 2: In a right triangle, each leg is the geometric mean of the hypotenuse
and the segment of the hypotenuse adjacent to it.
In the figure:
AD
AC
AC
AB
=
AD
BC
BC
AB
=
C
BA
D
C
B
D
A
C
A
B
D

7
Examples:
1. How long is the altitude of a right triangle that separates the hypotenuse
into lengths 4 and 20?
20
4 a
a
=
a2
= 80
a = 80
a = 516⋅
a = 54
2. Use the figure at the right to solve for x and y.
8
2 x
x
=
x2
= 16
x = 16
x = 4
Solve for y
8
6 y
y
=
y2
= 48
y = 48
y = 316⋅
y = 4 3
204
a
2
y
x
8

8
Try this out
A. Supply the missing parts:
1.
?
RT
RT
RW
=
2.
RS
TSWS
=
?
3.
RS
TS
TS
=
?
4.
?
RT
RT
RS
=
Give the indicated proportions.
5. The altitude is the geometric mean
6. The horizontal leg is the geometric mean
7. The vertical leg is the geometric mean
Find:
8. BS
9. RS
10.ST
B. Solve for x and y:
1.
x =
y =
F
S
OP
R
B
TS
8
2
x
4
7
y
T
S
W
R

9
x
y
2.
x =
y =
3.
x =
y =
4.
x =
y =
5.
x =
y =
C.
Given: Right ∆POM OR ⊥ PM,
xy
5
10
6
O
M
R
P
10
x
y
4 10
y
10
20
x

10
Find the missing parts:
1. PR = 5, RM = 10, OR =
2. OR = 6, RM = 9, PR =
3. PR = 4, PM =12, PO =
4. RM = 8, PM =12, OM =
5. PO = 9, PR = 3, PM =
6. PR = 6, RM = 8, PO =
7. PR = 4, RM = 12, OM =
8. PR = 4, PO = 6, RM =
9. PR = 8, OR = 12, RM =
10.PM =15, OM = 12, RM =
Lesson 3
The Pythagorean Theorem
In a right triangle, the square of the length of the hypotenuse is equal to the
sum of the squares of the lengths of the legs.
In the figure:
∆BCA is right with leg lengths,
a and b and hypotenuse length, c.
The Pythagorean Theorem in symbol:
c2
= a2
+ b2
Pythagorean Theorem is named after Pythagoras, a Greek Mathematician of
the sixth century BC. This theorem can be used to find a missing side length in a
right triangle.
Examples:
1. In the figure c = 13, b = 12
Find a:
c2
= a2
+ b2
132
= a2
+ 122
B
C A
a c
b
a = ?
c = 13
b = 12

11
a2
=132
- 122
a2
=169 – 144
a2
= 25
a = 25
a = 5
2. Find c, if a = 16 and b = 12
c2
= a2
+ b2
c2
= 162
+ 122
c2
= 256 + 144
c = 400
c = 20
3. c2
= a2
+ b2
82
= 42
+ b2
b2
= 82
– 42
b2
= 64 –16
b2
= 48
b = 48
b = 316⋅
b = 4 3
Try this out:
A. State whether the equation is correct or not
1.
a2
+ b2
= c2
2.
r2
= s2
+ t2
a = 16
b = 12
c = ?
a = 4
b = ?
c = 8
ac
b
rt
s

12
3.
e2
= f2
– d2
4.
a2
= c2
– b2
5.
k2
= l2
– m2
6.
p2
= r2
+ q2
7.
x2
= 32
+ 42
e
d
f
a
c
b
l
k
m
p
q
r
3x
4

13
8.
x2
= 102
– 72
9.
x2
= 42
+ 62
10.
x2
= 72
– 52
B. Write the equation you would use to find the value of x.
1.
2.
10
7
x
x4
6
x7
5
x4
3
x6
5

14
3.
4.
Classify each statement as true or false
5. 32
+ 42
= 52
6. 102
– 62
= 82
7. 12
+ 12
= 22
8. 22
+ 22
= 42
9. 72
– 52
= 52
10.92
+ 122
= 152
C. Given the lengths of two sides of a right triangle. Find the length of the third
side
1. a = 6, b = 8, c =
2. a = 5, b = 12, c =
3. a = 12, c = 15, b =
4. b = 4, c = 5, b =
5. a = 24, c = 26, b =
6. b = 16, c = 20
x
7
10
x5
6
ac
b
B
CA

15
7.
8.
9.
10
Lesson 4
Special Right Triangle
Isosceles Right Triangle or 45° – 45° – 90° Theorem:
In a 45° – 45° – 90° triangle, the length of the hypotenuse is equal to the
length of a leg times 2 .
In the figure:
If ∆ABC, a 45° – 45° – 90° triangle
when AC = BC = s then AB = s 2 .
30° - 60° - 90° Theorem:
In a 30° - 60° - 90° triangle, the length of the hypotenuse is twice the
length of the shorter leg, and the length of the longer leg is 3 times the length
of the shorter leg.
In the figure:
If ∆PRT where ∠R is a
right angle and ∠T = 30°,
Then:
a. PT = 2PR
b. RT = PR 3
a b c
7 24 ?
4 6 ?
7 9 ?
6 3 ? 12
ss
C
BA 45° 45°
P
RT
60°
30°

16
Examples:
1.Find the length of the hypotenuse of an isosceles
right triangle with a leg 7 2 cm long.
Hypotenuse = leg 2⋅ .
= 7 2 2⋅
= 7 · 2
= 14
2. Find the length of each leg of a 45° - 45° - 90°
triangle with a hypotenuse 12 cm long.
Leg =
2
hypotenuse
=
2
12
=
2
2
2
12
⋅ =
2
212
= 6 2 cm
3. Find the length of the longer leg and the length
of the hypotenuse.
Longer leg = shorter leg · 3
= 30 · 3
= 30 3 m
hypotenuse = shorter leg · 2
= 30 · 2
= 60 m
Try this out
A. Use the figure to answer the following:
1. The hypotenuse of a rt. ∆ABC is ___________.
2. The shorter leg of rt. ∆ABC is ___________.
7 2
45°45°
45°
45°
12 cm
30°
30 m
30°
30°
60°
60°
C
BA
D

17
3. The shorter leg of rt. ∆ADC is ___________.
4. The longest side of rt. ∆ADC is _________.
5. The altitude to the hypotenuse of ∆ACD is ________.
6. The longer leg of rt. ∆ACB is _________.
7. The longer leg of rt. ∆ADC is ________.
8. When CD = 2 then ____ = 4.
9. When CB = 6 then _____ = 6 3
10. When CB = 6 then _____ = 3
B. Find the value of x in each of the following:
1. 2. 3.
4. 5. 6.
7. 8. 9.
10.
x
45°
45°
10
30°
x
60°
24
30°
x
60°
30x
45°
45°
12
x
45°
45°
7
30°
x
60°
26
30°
x
60°
18
30°
60°
x
16
30°
60°
x
10
60°
30°
x
6

18
30°60°
x
y
10
C. Find the missing lengths, x and y.
1. 2.
45° 45°
3. 4.
5. 6.
7. 8.
60°
30°
9. 10.
45°
yx
5
45° 45°
x y
3
y
x
7
30°
10 3
x
y
30°
x
y
5
45°
45°
y
x
2
3
45° 45°
x y
2
3
45°
x
y
1.5
60°
30°
.5 3
x
y

19
Beyond the Pythagorean Theorem
In symbol c2
= a2
+ b2
, where c is the hypotenuse and a and b are the legs of
a right triangle.
Figure shows acute triangles
Figure shows obtuse triangles
Figure shown right triangle

20
Activity:
This activity will help you extend your understanding of the relationship of the
sides of a triangle.
Materials: Strips of paper cut in measured lengths of 2, 3, 4, 5, 6 and 8 units.
Procedure:
1. Form triangles with strips indicated by the number triplets below.
2. Draw the triangle formed for each number triple.
3. Fill out the table:
Number triplets
What kind of
triangle
Compute c2
Compute a2
+ b2
1. 3 4 5 Right 52
= 25 32
+ 42
= 25
2. 2 3 4
3. 2 4 5
4. 5 4 8
5. 6 5 8
6. 4 5 6
7. 2 3 3
8. 3 3 4
After the computation, the completed table will look like this
Number triplets
Kind of
triangle
c2
a2
+ b2 Comparison of c with
(a2
+ b2
)
1. 3 4 5 Right 25 25 Equal to
2. 2 3 4 Obtuse 16 13 Greater than
6. 4 5 6 Acute 36 41 Smaller than
1. What kind of ∆ did you get from triplet no. 1?
2. In triplet no. 1, what is the relation between c2
and (a2
+ b2
)?
3. Which triplets showed obtuse triangle?

21
4. For each obtuse triangle compare the result from c2
and (a2
+ b2
).
5. For acute triangles how will you compare the result of c2
and (a2
+ b2
)
Fill in the blanks with <, =, >:
6. In a right triangle, c2
____ a2
+ b2
7. In an obtuse triangle, c2
______ a2
+ b2
8. In an acute triangle, c2
____a2
+ b2
Let’s Summarize
Theorem: In a right triangle, the altitude to the hypotenuse separates the
triangle into two triangles each similar to the given triangle and
similar to each other.
Corollary 1: In a right triangle, the altitude to the hypotenuse is the
geometric mean of the segments into which it divides the
hypotenuse.
Corollary 2: In a right triangle, each leg is the geometric mean of the
hypotenuse and the segment of the hypotenuse adjacent to
it.
Pythagorean Theorem: The square of the length of the hypotenuse is
equal to the sum of the squares of the legs.
45°-45°-90° Theorem: In a 45°-45°-90° triangle, the length of the
hypotenuse is equal to the length of a leg times
2 .
30°- 60° - 90° Theorem: In a 30°- 60°- 90° triangle, the length of the
hypotenuse is twice the length of the shorter
leg, and the length of the leg is 3 times the

22
What have you Learned
Fill in the blanks:
1. The _______ to the hypotenuse of a right triangle forms two triangles each
similar to the given triangle & to each other.
2. The lengths of the ________ to the hypotenuse is the geometric mean of
the lengths of the segments of the hypotenuse.
3. In the figure
?
MA
MA
AB
=
for nos. 3 & 4
4. If BP = 8
AB = 4
Find PM ___
5. If in a right triangle the lengths of the legs are 8 and 15, the length of the
hypotenuse is _______
6. Find the length of an altitude of an equilateral triangle if the length of a
side is 10.
7. In a 30° – 60° – 90° triangle, the length of the hypotenuse is 8. Find the
8. - 9. ∆ACB is an isosceles right triangle. CD bisects ∠C, the right angle.
Find AB and CB.
10. What is the height of the Flag Pole?
M
P
B
A
C
A B
D
3
2 m
8 m

23
Answer key
How much do you know
1. AB 6. 5
4
3
2. ∆ACB 7. 8 2
3. CD 8. 3 3
4. 10 9. 10 2
5. 7 3 10. 2 21
Lesson 1:
A. B.
1. ∠C or ∠ACB 1. ∆ROS ∼ ∆RST
2. CD 2. ∆TOS ∼ ∆RST
3. AB 3. ∆ROS ∼ ∆TOS
4. BD or DB
5. BC 4. ∆MST ∼ ∆MOR
6. ∠ADC 5. ∆RSO ∼ ∆MOR
7. BC 6. ∆MST ∼ ∆RSO
8. ∆BDC
9. ∆ADC
10.∆BDC
C.
1. ∆ADC, ∆BDC, ∆ACB
2. CD
3. AC
4. AB
5. BD
6. GH
7. OK
8. ∆MNR ∼ ∆MPO
9. ∆ONP ∼ ∆MPO
10. ∆MNP ∼ ∆ONP
Lesson 2
A. B. C.
1. WS 1. x = 2 11 1. 5 2
2. TS y = 2 7 2. 4

24
3. WS 2. x = 10 3 3. 4 3
4. RW y = 30 4. 4 6
5.
SP
OS
OS
SF
= 3. x = 2 14 5. 27
y = 2 35 6. 4 3
6.
PF
PO
PO
SP
= 4. x = 20 7. 8 3
y = 5 5 8. 9
7.
PF
OF
OF
FS
= 5. x =
3
2
16
6
100
or 9. 18
y = 10 10. 9.6
8. 4
9. 2 5
10. 4 5
Lesson 3
A. B. C.
1. correct 1. x2
= 32
+ 42
1. 10
2. correct 2. x2
= 62
– 52
2. 13
3. not 3. x2
= 102
– 72
3. 9
4. correct 4. x2
= 62
– 52
4. 3
5. correct 5. true 5. 10
6. not 6. true 6. 12
7. correct 7. false 7. 25
8. correct 8. false 8. 2 13
9. not 9. false 9. 130
10. not 10. true 10. 6
Lesson 4
A. B. C. x y
1. AB 1. 12 1. 5 3 5
2. BD 2. 10 3 2.
2
3
2
23
3. CD 3. 8 3. 3 3
4. AC 4. 18 3 4.
2
23
2
23

25
5. CD 5. 13 5.
2
25
2
25
6. AC 6. 7 2 6.
2
25.1
2
25.1
7. AD 7. 12 3 7. 14 7 3
8. AC 8. 12 2 8. 5 5 3
9. AC 9. 15 3 9. 5 3 15
10. DB 10.
3
310
10. 5 10
What have you learned
1. altitude
2. altitude
3. AP
4. 4 2
5. 17
6. 5 3
7. 4
8. 3 2
9. 3
10. 2 3

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