1. Sec. 11.3 Perimeter and Area
of Similar Figures p. 737
(You will need your workbook p. 211.)
Objective: To use ratios to find
areas of similar figures
Vocabulary: regular polygon, p.
43, corresponding sides, p. 225,
similar polygons, p. 372

2. Two polygons are
similar polygons if
and only if their
corresponding angles
are congruent and their
corresponding side
lengths are
proportional.
Let’s review similar figures.
Also, remember that all lengths in
similar figures have the same ratio.

3. Introduction to Lesson:
One TV screen is 32 in. x 20 in. and
another is 48 in. x 30 in. How would you
show whether the screens are similar?
If , then they are similar.
Now, calculate the area of each,
(640 in.2 and 1440 in.2), the ratio of sides,
(2/3), and the ratio of areas.
What did you discover?
48
32
= 30
20
1440
640
= 9
4

4. EXAMPLE 1 Find ratios of similar polygons
Ratio (red to blue) of the
perimeters
a.
Ratio (red to blue) of the
areas
b.
In the diagram, ABC DEF. Find the indicated ratio.∆ ∆
The ratio of the lengths of corresponding sides is
12
8
= 3
2 , or 2:3.
SOLUTION
a. The ratio of the perimeters is also 2:3.
b. The ratio of the areas is 22:32, or 4:9.

5. The perimeter of ∆ABC is 16 feet, and its area is
64 square feet. The perimeter of ∆DEF is 12 feet.
Given ∆ABC ~ ∆DEF, find the ratio of the area of ∆ABC
to the area of ∆DEF. Then find the area of ∆DEF.
GUIDED PRACTICE for Examples 1 and 2
The ratio of the lengths of corresponding perimeters is
12
16
= 3
4 , or 4:3.
The ratio of corresponding area is .
9
16
So,…
9
16
= x
64 and x = 36 ft2

6. EXAMPLE 3 Use a ratio of areas
If the area ratio is a2:b2, then the length ratio is a:b.
Cooking
A large rectangular baking pan is 15 inches long and 10 inches wide. A
smaller pan is similar to the large pan. The area of the smaller pan is 96
square inches. Find the width of the smaller pan.
SOLUTION
Area of smaller pan
Area of large pan =
150
96 =
25
16
Length in smaller pan
Length in large pan =
4
5
Any length in the smaller pan is , or 0.8, of the corresponding length in
the large pan. So, the width of the smaller pan is 0.8(10 inches) 8 inches.
4
5
=

7. EXAMPLE 2 Standardized Test Practice
SOLUTION
The ratio of a side length of the den to the
corresponding side length of the bedroom is 14:10, or
7:5. So, the ratio of the areas is 72:52, or 49:25. This ratio
is also the ratio of the carpeting costs. Let x be the cost
for the den.

8. EXAMPLE 2 Standardized Test Practice
25
49 =
225
x cost of carpet for den
cost of carpet for bedroom
x 441= Solve for x.
ANSWER
It costs $441 to carpet the den. The correct answer is
D.