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Section 10–4 Perimeters & Areas of Similar Figures

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Section 10–4 Perimeters & Areas of Similar Figures

Objectives:

1) To find perimeters & areas of similar figures.

- Perimeter – Distance around a figure
- Perimeter of any polygon - add up the lengths of all of the sides
- Perimeter of a circle – Circumference
- C = 2r

- Area – How much 2D space it takes up
- A// = bh
- AΔ = ½ bh
- A = r2

- If the similarity (side) ratio of 2 similar figures is a/b, then
- The ratio of their perimeters is a/b.
- The ratio of their areas is a2/b2.

b

a

ΔABC ~ ΔFDE

D

5

Side ratio =

4

6.25

5

B

E

F

7.5

6

Perimeter Ratio = Side Ratio

Perimeter Ratio = 5/4

Area Ratio = a2/b2 =

4

A

C

5

52/42 =

25/16

The ratio of the lengths of the corresponding sides of 2 regular octagons is 8/3. The area of the larger octagon is 320ft2. Find the area of the smaller octagon.

8

Side ratio =

3

Now, set up an area proportion using the area ratio!

82

64

Area ratio =

=

32

9

Large side

Large Area

64

320

=

9

x

x = 45ft2

Small side

The areas of 2 similar pentagons are 32in2 and 72in2. What is their similarity (side) ratio? What is the ratio of their perimeter.

Reduce

Remember: Side ratio is a/b and area ratio is a2/b2. So if the area ratio is given, you must take the square root of the numerator and the denominator.

32

4

2

=

=

3

72

9

Area Ratio

Side Ratio and the Perimeter ratio

The similarity (side) ratio of two similar Δis 5:3. The perimeter of the smaller Δ is 36cm, and its area is 18cm2. Find the perimeter & area of the larger Δ.

Write the side ratio and then find the perimeter.

Write the area ratio and then find the area.

52

25

A

5

P

=

=

=

32

9

18

3

36

PL = 60cm

A = 50cm2

- Side Ratio = a/b
- Perimeter Ratio = a/b
- Area Ratio = a2/b2
- If perimeters are given:
- Write as a ratio
- Reduce to simplest form for the side ratio

- If Areas are given:
- Write as a ratio
- Reduce until 2 perfect squares are reached.
- Square Root (√) both numerator & denominator for the side ratio