Post on 31-Dec-2015
description
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
π=54π΄=162
π΄=72π=36
69=
23
3654
=46=
23
72162
=3681
=49
π2=25+16π=β41
π=9+β 41π΄=10
π2=100+64π=2β 41π=18+2β41
π΄=40
12
9+β 4118+2β41
1040
=14
(8)
πΆ=16ππ΄=64π
(6)
πΆ=12ππ΄=36π
68=
34
12π16π
ΒΏ34
36π64π
ΒΏ9
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=
36π₯ 3 π₯=180 π₯=60ππ
925
=π₯
80 25 π₯=720 π₯=28.8ππ2