# Ratio and Proportion - PowerPoint PPT Presentation

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Ratio and Proportion. 7-1. 5 ft. 5 ft. 5 ft. 20 in . 20 in . 20 in . b. . a. . 64 m : 6 m. 64 m. 64 m. Write. 64 m : 6 m as. . 6 m. 6 m. a. . 3. 32. 1. 3. =. =. 32 : 3. To simplify a ratio with unlike units, multiply by a conversion factor. b.

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Ratio and Proportion

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## Ratio and Proportion

7-1

5 ft

5 ft

5 ft

20 in.

20 in.

20 in.

b.

a.

64 m : 6 m

64 m

64 m

Write

64 m : 6 mas

.

6 m

6 m

a.

3

32

1

3

=

=

32 : 3

To simplify a ratio with unlike units, multiply by a conversion factor.

b.

60

12 in.

20

1 ft

=

=

=

EXAMPLE 1

Simplify ratios

Simplify the ratio.

SOLUTION

Then divide out the units and simplify.

24

24

3

3

Write

24 yards : 3 yards as

8

1

=

=

8 : 1

for Example 1

GUIDED PRACTICE

Simplify the ratio.

1. 24 yardsto 3 yards

SOLUTION

Then divide out the units and simplify.

150

6

1

4

150cm

1m

=

=

6m

100cm

for Example 1

GUIDED PRACTICE

Simplify the ratio.

2. 150 cm : 6 m

SOLUTION

To simplify a ratio with unlike units, multiply by a conversion factors.

= 1 : 4

x

5

x

5

a.

=

16

10

16

10

Solve the proportion.

ALGEBRA

a.

=

5 16

=

80

10 x

=

10 x

8

x

=

EXAMPLE 4

Solve proportions

SOLUTION

Write original proportion.

Cross Products Property

Multiply.

Divide each side by 10.

1

1

2

2

b.

=

y + 1

y + 1

3y

3y

b.

=

1 3y

2 (y + 1)

=

3y

2y + 2

=

y

2

=

EXAMPLE 4

Solve proportions

SOLUTION

Write original proportion.

Cross Products Property

Distributive Property

Subtract 2y from each side.

2

2

5

5

5.

=

x

8

8

x

=

2 8

5 x

=

16

5 x

=

16

x

=

5

for Example 4

GUIDED PRACTICE

SOLUTION

Write original proportion.

Cross Products Property

Multiply.

Divide each side by 5 .

1

1

4

4

6.

=

x – 3

3x

x – 3

3x

=

3x

=

4(x – 3)

3x

=

4x – 12

3x – 4x

=

– 12

– x

x

– 12

12

=

=

for Example 4

GUIDED PRACTICE

SOLUTION

Write original proportion.

Cross Products Property

Multiply.

Subtract 4x from each side.

y – 3

y

y

7.

=

=

7

14

14

y – 3

7

=

7 y

14(y – 3)

14y – 42

=

7y

14y –7y

=

42

y

6

=

for Example 4

GUIDED PRACTICE

SOLUTION

Write original proportion.

Cross Products Property

Multiply.

Subtract 7y from each side and add 42 to each side.

Subtract , then divide

Painting

You are planning to paint a mural on a rectangular wall. You know that the perimeter of the wall is 484feet and that the ratio of its length to its width is 9 : 2. Find the area of the wall.

STEP 1

Write expressions for the length and width. Because the ratio of length to width is 9 : 2, you can represent the length by 9xand the width by 2x.

EXAMPLE 2

Use a ratio to find a dimension

SOLUTION

STEP 2

Solve an equation to findx.

2l + 2w

=

P

484

2(9x) + 2(2x)

=

484

22x

=

x

22

=

Evaluate the expressions for the length and width. Substitute the value of xinto each expression.

STEP 3

The wall is 198feet long and 44feet wide, so its area is

198 ft

44 ft

=

8712 ft.

2

EXAMPLE 2

Use a ratio to find a dimension

Formula for perimeter of rectangle

Substitute for l, w, and P.

Multiply and combine like terms.

Divide each side by 22.

Length= 9x = 9(22) = 198

Width = 2x = 2(22) = 44

ALGEBRA The measures of the angles in CDE are in the extended ratio of 1 : 2 : 3. Find the measures of the angles.

o

o

o

o

180

x + 2x + 3x

=

6x

180

=

x

=

30

o

o

o

o

o

The angle measures are 30 , 2(30 ) = 60 , and 3(30 ) = 90.

EXAMPLE 3

Use extended ratios

SOLUTION

Begin by sketching the triangle. Then use the extended ratio of 1 : 2 : 3 to label the measures as x° , 2x° , and 3x° .

Triangle Sum Theorem

Combine liketerms.

Divide each side by 6.

STEP 1

Write expressions for the length and width. Because the ratio of length is 7 : 5, you can represent the length by 7xand the width by 5x.

for Examples 2 and 3

GUIDED PRACTICE

3. The perimeter of a room is 48 feet and the ratio of its length to its width is 7 : 5. Find the length and width of the room.

SOLUTION

STEP 2

Solve an equation to findx.

2l + 2w

=

P

48

2(7x) + 2(5x)

=

48

24x

=

x

2

=

Evaluate the expressions for the length and width. Substitute the value of xinto each expression.

STEP 3

for Examples 2 and 3

GUIDED PRACTICE

Formula for perimeter of rectangle

Substitute for l, w, and P.

Multiply and combine like terms.

Length= 7x + 7(2) = 14 ft

Width = 5x + 5(2) = 10 ft

4.A triangle’s angle measures are in the extended ratio of 1 : 3 : 5. Find the measures of the angles.

x

3x 5x

o

o

o

o

180

x + 3x + 5x

=

9x

180

=

x

=

20

o

o

o

o

o

The angle measures are 20 , 3(20 ) = 60 , and 5(20 ) = 100.

for Examples 2 and 3

GUIDED PRACTICE

SOLUTION

Begin by sketching the triangle. Then use the extended ratio of 1 : 3 : 5 to label the measures as x° , 2x° , and 3x° .

Triangle Sum Theorem

Combine liketerms.

Divide each side by 9.