EXAMPLE 3
Use a ratio of areas
Cooking
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.
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.
EXAMPLE 3
Use a ratio of areas
Area of smaller pan = 96
150
Area of large pan
Length in smaller pan = 4
5
Length in large pan
Write ratio of known
= 16
25 areas. Then simplify.
Find square root of
area ratio.
ANSWER
4
Any length in the smaller pan is 5 , 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.
EXAMPLE 4
Solve a multi-step problem
Gazebo
The floor of the gazebo shown is a
regular octagon. Each side of the
floor is 8 feet, and the area is about
309 square feet.You build a small
model gazebo in the shape of a
regular octagon. The perimeter of
the floor of the model gazebo is 24
inches. Find the area of the floor of
the model gazebo to the nearest
tenth of a square inch.
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.
STEP 1 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.
Perimeter of full-sized = 8(8 ft) = 64 ft = 32
1
24 in.
2 ft
Perimeter of model
So, the ratio of corresponding lengths (full-sized to
model) is 32:1.
EXAMPLE 4
Solve a multi-step problem
STEP 2 Calculate the area of the model gazebo’s floor.
Let x be this area.
(length of full-sized)2
(length of model)2
=
Area of full-sized
Area of model
322 = 309 ft2
x ft2
12
1024x = 309
x ≈ 0.302 ft2
Theorem
11.7
Substitute.
Cross Products
Property
Solve for x.
EXAMPLE 4
Solve a multi-step problem
STEP 3 Convert the area to square inches.
0.302 ft2 144 in.2 ≈ 43.5 in.2
1 ft.2
ANSWER
The area of the floor of the model gazebo is about
43.5 square inches.
GUIDED PRACTICE
2.
for Examples 3 and 4
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.
Area of smaller decagon
Area of large decagon
Write ratio of
= 20 = 59 known areas.
36
Length in smaller decagon = √ 5
3
Length in large decagon
Then simplify.
Find square root
of area as
ratios.
GUIDED PRACTICE
3.
for Examples 3 and 4
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)
= 2
35 + 2
= 110 feet.
20
GUIDED PRACTICE
for Examples 3 and 4
Convert 110 feet to inches.
110 ft = 110
12 = 1320 inches.
STEP 1 Find the ratio of the lengths of the two
rectangles by finding the ratio of the
perimeters.
Perimeter of Rectangle I = 66 = 1
20
1320
Perimeter of Rectangle II
So, the ratio of the corresponding lengths to is 1:20.
GUIDED PRACTICE
for Examples 3 and 4
STEP 2 Find the area of Rectangle II.
Area of Rectangle II = lb
= 35(20)
= 700 feet2
Convert 700 feet2 to inches2.
700 ft2 = 700
144 = 100,800 inches2.
STEP 3 Find the ratio of their Areas
(sides of Rectangle I)2 = Area of Rectangle I
(sides of Rectangle II)2
Area of Rectangle II
1 = 1
202
400
GUIDED PRACTICE
for Examples 3 and 4
STEP 4 Find the area of Rectangle I.
Area of Rectangle I = Area of Rectangle II · (ratio of areas)
1
Area of Rectangle I = 100,800 · 400
= 252 in.2