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A large rectangular baking pan is 15 inches long and 10 inches wide. A smaller pan is similar to the large pan. The area of the smaller pan is 96 square inches . Find the width of the smaller pan. EXAMPLE 3. Use a ratio of areas. Cooking. SOLUTION.

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A large rectangular baking pan is 15 inches long and 10inches wide. A smaller pan is similar to the large pan. The area of the smaller pan is 96square inches. Find the width of the smaller pan.

EXAMPLE 3

Use a ratio of areas

Cooking

SOLUTION

First draw a diagram to represent the problem. Label dimensions and areas.

Then use Theorem 11.7. If the area ratio is a2:b2, then the length ratio is a:b.

Area of smaller pan

96

16

=

=

25

150

Area of large pan

4

Length in smaller pan

=

5

Length in large pan

4

Any length in the smaller pan is , or 0.8, of the corresponding length in the large pan. So, the width of the smaller pan is 0.8(10 inches)8 inches.

5

=

EXAMPLE 3

Use a ratio of areas

Write ratio of known areas. Then simplify.

Find square root of area ratio.

Gazebo

EXAMPLE 4

Solve a multi-step problem

The floor of the gazebo shown is a regular octagon. Each side of the floor is 8 feet, and the area is about 309square feet.You build a small model gazebo in the shape of a regular octagon. The perimeter of the floor of the model gazebo is 24inches. Find the area of the floor of the model gazebo to the nearest tenth of a square inch.

64 ft

Perimeter of full-sized

32

8(8 ft)

=

=

=

1

24 in.

Perimeter of model

2 ft

EXAMPLE 4

Solve a multi-step problem

SOLUTION

All regular octagons are similar, so the floor of the model is similar to the floor of the full-sized gazebo.

Find the ratio of the lengths of the two floors by finding the ratio of the perimeters. Use the same units for both lengths in the ratio.

STEP 1

So, the ratio of corresponding lengths (full-sized to model) is 32:1.

Area of full-sized

(length of full-sized)2

=

Area of model

(length of model)2

322

309 ft2

=

x ft2

12

1024x 309

=

EXAMPLE 4

Solve a multi-step problem

STEP 2

Calculate the area of the model gazebo’s floor. Let xbe this area.

Theorem11.7

Substitute.

Cross Products Property

x ≈0.302 ft2

Solve for x.

0.302

ft2

144in.2

≈ 43.5in.2

1ft.2

The area of the floor of the model gazebo is about

43.5square inches.

EXAMPLE 4

Solve a multi-step problem

STEP 3

Convert the area to square inches.

Area of smaller decagon

20

5

=

=

9

36

Area of large decagon

5

Length in smaller decagon

=

3

Length in large decagon

for Examples 3 and 4

GUIDED PRACTICE

2. The ratio of the areas of two regular decagons is

20:36. What is the ratio of their corresponding side lengths in simplest radical form?

SOLUTION

Then use Theorem 11.7. If the area ratio is a2:b2, then the length ratio is a:b.

Write ratio of known areas. Then simplify.

Find square root of area as ratios.

= 2 35 + 2 20

for Examples 3 and 4

GUIDED PRACTICE

3. Rectangles I and II are similar. The perimeter of

Rectangle I is 66 inches. Rectangle II is 35 feet long and 20 feet wide. Show the steps you would use to

find the ratio of the areas and then find the area of

Rectangle I.

SOLUTION

Perimeter of Rectangle I = 66 inches

Perimeter of Rectangle II = 2(l + b)

= 2(35 + 20)

= 110 feet.

= 110 12

1

66

Perimeter of Rectangle I

=

=

20

1320

Perimeter of Rectangle II

for Examples 3 and 4

GUIDED PRACTICE

Convert110 feet to inches.

110 ft

= 1320 inches.

Find the ratio of the lengths of the two rectangles by finding the ratio of the perimeters.

STEP 1

So, the ratio of the corresponding lengths to is 1:20.

= 700 144

(sides of RectangleI)2

Area of RectangleI

=

(sides of RectangleII)2

Area of RectangleII

1

1

=

202

400

for Examples 3 and 4

GUIDED PRACTICE

STEP 2

Find the area of Rectangle II.

Area of Rectangle II = lb

= 35(20)

= 700 feet2

Convert700 feet2to inches2.

700 ft2

= 100,800 inches2.

STEP 3

Find the ratio of their Areas

Area of Rectangle I = 100,800 ·

1

400

for Examples 3 and 4

GUIDED PRACTICE

STEP 4

Find the area of Rectangle I.

Area of Rectangle I = Area of Rectangle II · (ratio of areas)

= 252 in.2