10.4 Perimeters and Areas of Similar

Polygons

Geometry

Objectives: SWBAT compare perimeters and areas of

similar figures. SWBAT use perimeters and areas of similar

figures to solve real-life problems.

DFA: pp.555-556 #4 & 6

HW: pp. 555-557 (2-40 even)

Objectives/DFA/HW

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.

Comparing Perimeter and Area

In lesson 8.3, 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. In activity 11.3 on pg. 676, you may have discovered that the ratio of the areas of two similar polygons is NOT this same ratio.

Comparing Perimeter and Area

Theorem 10.7 – Areas of Similar Polygons

Thm 11.5 continuedSide length of Quad I

Side length of Quad II=

a

b

Area of Quad I

Area of Quad II=

a2

b2

I II

kakb

Quad I ~ Quad II

Ex. 1: Finding Ratios of Similar Polygons

Pentagons ABCDE and LMNPQ are similar.

a. Find the ratio (red to blue) of the perimeters of the pentagons.

b. Find the ratio (red to blue) of the areas of the pentagons

A

B

C D

E

L

M

N P

Q

5

10

Ex. 1: Solutiona. 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.

A

B

C D

E

L

M

N P

Q

5

10

Ex. 1: Solutionb. Find the ratio (red to blue) of

the areas of the pentagons. Using Theorem 10.7, 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.

A

B

C D

E

L

M

N P

Q

5

10

Using perimeter and area in real life

Ex. 2: Using Areas of Similar Figures

Because the ratio of the lengths of the sides of to rectangular pieces is 1:2, the ratio of the areas of the pieces of paper is 12: 22 or, 1:4

Using perimeter and area in real life

Ex. 2: Using Areas of Similar Figures

Because the cost of the paper should be a function of its area, the larger piece of paper should cost about 4 times as much, or $1.68.

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 14.25 feet and an area of about 980.4 square feet. Find the area of a smaller octagon that has a perimeter of about 76 feet.

Using perimeter and area in real life

Ex. 3: Finding Perimeters and Areas of Similar Polygons

Using perimeter and area in real life

All regular octagons are similar because all corresponding angles are congruent and corresponding side lengths are proportional.

First – Draw and label a sketch.

Ex. 3: Solution

FIND the ratio of the side lengths of the two octagons, which is the same as the ratio of their perimeters.

Using perimeter and area in real life

Ex. 3: Solution

perimeter of ABCDEFGH

perimeter of JKLMNPQR=

a

b

76

8(14.25)=

76

114=

2

3

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.

Using perimeter and area in real life

Ex. 3: Solution

9

A

980.4=

4

9

9A = 980.4 • 4

A = 3921.6

A 435.7

Write the proportion.

Cross product property.

Divide each side by 9.

Use a calculator.

The area of the smaller octagon is about 435.7 square feet.