AREAS OF SIMILAR POLYGONS
THEOREM
THEOREM 11.5 Areas of Similar Polygons
If two polygons are similar with the lengths
of corresponding sides in the ratio of a:b,
then the ratio of their areas is a2:b2.
I
II
11.3
Side length of Quad. I
a
Side length of Quad. II = b
Area of Quad. I
Area of Quad. II
Perimeters and Areas of Similar Figures
a2
= b2
Finding Ratios of Similar Polygons
Find the ratio (red to blue) of the perimeters and of the areas.
SOLUTION
3
3
9
=
1
3
12 = 1
32
9
11.3
Perimeter = 1:3
Area = 12 : 32 = 1:9
Perimeters and Areas of Similar Figures
9
Finding Ratios of Similar Polygons
Find the ratio (red to blue) of the perimeters and of the areas.
SOLUTION
5
12.5
3
7.5
Perimeter = 5:3
Area = 52 : 32 = 25:9
11.3
Perimeters and Areas of Similar Figures
Finding Ratios of Similar Polygons
Find the ratio of the side lengths of the two octagons, which is
the same as the ratio of their perimeters.
SOLUTION
ABCDEFGH
perimeter ≈ 76 ft
All regular octagons are similar
because all corresponding s are
.
JKLMNPQR
side length ≈ 14.25 ft
area ≈ 980.4 ft2
perimeter ABCDEFGH = a ≈
76
= 76 = 2
perimeter JKLMNPQR
b
8(14.25)
114
3
11.3
Perimeters and Areas of Similar Figures
Finding Ratios of Similar Polygons
Calculate the area of the smaller octagon.
SOLUTION
ABCDEFGH
perimeter ≈ 76 ft
Let A represent the area of the smaller
octagon. The ratio of the the areas of the
smaller to the larger is a2:b2 = 22:32, or 4:9.
A
= 4
980.4
9
Write a proportion.
9A = 980.4 x 4
Cross product property
A = 3921.6
9
Division property of =
11.3
Perimeters and Areas of Similar Figures
JKLMNPQR
area ≈ 980.4 ft2
Finding Ratios of Similar Polygons
Calculate the area of the smaller octagon.
SOLUTION
ABCDEFGH
perimeter ≈ 76 ft
A
= 4
980.4
9
Write a proportion.
9A = 980.4 x 4
Cross product property
A = 3921.6
9
Division property of =
A ≈ 435.7
JKLMNPQR
area ≈ 980.4 ft2
Solve
 The area of the smaller octagon is about 435.7 ft2
11.3
Perimeters and Areas of Similar Figures