NAME
DATE
11-5
PERIOD
Study Guide and Intervention
Areas of Similar Figures
Areas of Similar Figures
If two polygons are similar, then their areas are
proportional to the square of the scale factor between them.
Example
JKL ∼ PQR.
The area of JKL is 40 square inches.
Find the area of PQR.
JJ
6
12
or −
.
Find the scale factor: −
10
( )
R
10in.
in.
10
5
6 2
The ratio of their areas is − .
5
area of PQR
6
− = −
5
area ofJKL
R
( )
L
K
Write a proportion.
area of PQR
36
− =−
40
25
36
area of PQR = −
40
25
L
K
2
( 5)
6
Area of JKL = 40; −
2
36
=−
P
P
12 in.
Q
12 in.
Q
C11-019A-890520-A-A
25
C11-019A-890520-A
Multiply each side by 40.
area of PQR = 57.6
Simplify.
So the area of PQR is 57.6 square inches.
For each pair of similar figures, find the area of the shaded figure. Round to the
nearest tenth if necessary.
2.
1.
5 m
15 m
2 in.
6 in.
A = 20 in2
A = 12 m2
4.
A = 200 cm2
Lesson 11-5
A = 8050 ft2
435.8 cm2
5152 ft2
C11-019A-890520-D
Chapter 11
ft
10.5 cm
C11-019A-890520-C
16
3.
C11-019A-890520-B
15.5 cm
180 in2
ft
108 m2
20
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
Exercises
C11-019A-890520-E
31
Glencoe Geometry
NAME
DATE
11-5
PERIOD
Study Guide and Intervention (continued)
Areas of Similar Figures
Scale Factors and Missing Measures in Similar Figures You can use the areas
of similar figures to find the scale factor between them or a missing measure.
Example
A
If ABDC is similar to
FGJH, find the value of x.
Let k be the scale factor between ABDC and FGJH.
area ABCD
− = k2
area FGJH
64
−
= k2
49
8
−
=k
7
B F
C
D H
10 m
A = 64 m2
Theorem 11.1
Substitution
G
x
J
A = 49 m2
C11-020A-890520-A
Take the positive square root of each side.
Use this scale factor to find the value of x.
CD
−
=k
The ratio of corresponding lengths of similar polygons is equal to the scale factor between the
HJ
10
8
−
x =−
7
7
x = − 10 or 8.75
8
polygons.
Substitution
Multiply each side by 10.
For each pair of similar figures, use the given areas to find the scale factor from
the unshaded to the shaded figure. Then find x.
2.
1.
x ft
8
x
in
.
28 ft
A = 296 ft2
A = 54 in2
A = 216
A = 169 ft2
in2
√ 169
296
−
k = C11-020A-890520-C
; x = 21.2 ft
1
k=−
; x = 16 in.
2C11-020A-890520-B
3.
4.
7 ft
x cm
A = 300 cm2
21 cm
A = 900 cm2
A = 50 ft2
3 ; x = 12.1
k = √
k=
C11-020A-890520-D
Chapter 11
x
32
A = 30 ft2
5
; x = 5.4 ft
√−C11-020A-890520-E
3
Glencoe Geometry
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
Exercises