1. Perimeters and Areas of Similar Polygons

2. Comparing Perimeter and Area
For any polygon, the perimeter of the polygon is the sum of the lengths of its sides and the area of the polygon is the number of square units contained in its interior.

3. Comparing Perimeter and Area
In chapter 7 before, you learned that if two polygons are similar, then the ratio of their perimeters is the same as the ratio of the lengths of their corresponding sides.

4. 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

5. Quad I ~ Quad II kb Side length of Quad I a = ka
Side length of Quad II b I II
Area of Quad I a2 =
Area of Quad II b2
Quad I ~ Quad II

6. Ex. 1: Finding Ratios of Similar Polygons
Pentagons ABCDE and LMNPQ are similar.
Find the ratio (red to blue) of the perimeters of the pentagons.
Find the ratio (red to blue) of the areas of the pentagons 5 10

7. Ex. 1: Solution
Find the ratio (red to blue) of the perimeters of the pentagons.
The ratios of the lengths of corresponding sides in the pentagons is 5:10 or ½ or 1:2.
The ratio is 1:2. So, the perimeter of pentagon ABCDE is half the perimeter of pentagon LMNPQ. 5 10

8. Ex. 1: Solution
Find the ratio (red to blue) of the areas of the pentagons.
Using Theorem 11.5, the ratio of the areas is 12: 22. Or, 1:4. So, the area of pentagon ABCDE is one fourth the area of pentagon LMNPQ. 5 10

9. Using perimeter and area in real life
Ex. 2: Finding Perimeters and Areas of Similar Polygons
Octagonal Floors. A trading pit at the Chicago Board of Trade is in the shape of a series of octagons. One octagon has a side length of about feet and an area of about square feet. Find the area of a smaller octagon that has a perimeter of about 76 feet.

10. Using perimeter and area in real life
Ex. 2: Solution
All regular octagons are similar because all corresponding angles are congruent and corresponding side lengths are proportional.
First – Draw and label a sketch.

11. Using perimeter and area in real life
Ex. 2: Solution
FIND the ratio of the side lengths of the two octagons, which is the same as the ratio of their perimeters. a 76 76 2
perimeter of ABCDEFGH =  = = b
8(14.25)
114 3
perimeter of JKLMNPQR

12. Using perimeter and area in real life
Ex. 2: Solution
CALCULATE the area of the smaller octagon. Let A represent the area of the smaller octagon. The ratio of the areas of the smaller octagon to the larger is a2:b2 = 22:32, or 4:9. A
980.4 = 4 9
Write the proportion.
The area of the smaller octagon is about square feet.
9A = • 4
Cross product property.
A =
Divide each side by 9. 9
Use a calculator.
A  435.7