Geometry

Unit VIIPerimeters and Areas of Similar

Figures

Objective: to compare and solve perimeters and areas of similar figures

two rectangles two right triangles two circles

Scale __________ __________ __________ factor Ratio of __________ __________ __________ perimeters Ratio of __________ __________ __________ the areas

8

6

5

4 10

8

9

6

18

12

𝑃=54𝐴=162

𝐴=72𝑃=36

69=

23

3654

=46=

23

72162

=3681

=49

𝑐2=25+16𝑐=√41

𝑃=9+√ 41𝐴=10

𝑐2=100+64𝑐=2√ 41𝑃=18+2√41

𝐴=40

12

9+√ 4118+2√41

1040

=14

(8)

𝐶=16𝜋𝐴=64𝜋

(6)

𝐶=12𝜋𝐴=36𝜋

68=

34

12𝜋16𝜋

¿34

36𝜋64𝜋

¿9

16

Perimeters of Similar Polygons: Two similar figures with lengths of corresponding sides in the ratio of

a:b will have perimeters in the ratio of a:b.

Theorem 11.5: Areas of similar polygons: Two similar figures with lengths of corresponding sides in the

ratio of a:b will have areas in the ratio of .

Examples: The ratio of corresponding sides in two similar polygons is 3:5. a) Find the ratio of the perimeters. b) Find the ratio of the areas. c) If the perimeter of the first polygon is 36cm, find the perimeter of the second polygon. d) If the second polygon has an area of 80cm2, find the area of the first polygon.

3 :5

9 :25

35=

36𝑥 3 𝑥=180 𝑥=60𝑐𝑚

925

=𝑥

80 25 𝑥=720 𝑥=28.8𝑐𝑚2