Ratio and Proportion 7-1. EXAMPLE 1 Simplify ratios SOLUTION 64 m : 6 m a. Then divide out the units...
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Transcript of Ratio and Proportion 7-1. EXAMPLE 1 Simplify ratios SOLUTION 64 m : 6 m a. Then divide out the units...
Ratio and ProportionRatio and ProportionRatio and ProportionRatio and Proportion
7-17-1
EXAMPLE 1 Simplify ratios
SOLUTION
64 m : 6 ma.
Then divide out the units and simplify.
b. 5 ft20 in.
b. To simplify a ratio with unlike units, multiply by a conversion factor.
a. Write 64 m : 6 m as 64 m6 m
.
= 323
= 32 : 3
5 ft20 in. = 60
20 = 31= 5 ft
20 in.12 in.1 ft
Simplify the ratio.
64 m6 m
GUIDED PRACTICE for Example 1
SOLUTION
Then divide out the units and simplify.
24 3
= 81
= 8 : 1
Simplify the ratio.
1. 24 yards to 3 yards
Write 24 yards : 3 yards as 24 3
GUIDED PRACTICE for Example 1
2. 150 cm : 6 m
SOLUTION
To simplify a ratio with unlike units, multiply by a conversion factors.
= 14 = 1 : 4
Simplify the ratio.
150cm 6m = 150
6 1m100cm
EXAMPLE 4 Solve proportions
SOLUTION
a. 510
x16=
Multiply.
Divide each side by 10.
a. 510
x16=
= 10 x5 16
= 10 x80
= x8
Write original proportion.
Cross Products Property
Solve the proportion.ALGEBRA
EXAMPLE 4 Solve proportions
Subtract 2y from each side.
1y + 1
23y
b. =
= 2 (y + 1)1 3y
= 2y + 23y
=y 2
Distributive Property
SOLUTION
b. 1y + 1 = 2
3y
Write original proportion.
Cross Products Property
GUIDED PRACTICE for Example 4
5. 2 x
5 8=
SOLUTION
Write original proportion.
Cross Products Property
Multiply.
Divide each side by 5 .
2 x
5 8=
= 5 x2 8
= 5 x16
=x 16 5
GUIDED PRACTICE for Example 4
6. 1x – 3
43x=
SOLUTION
Write original proportion.
Cross Products Property
Multiply.
Subtract 4x from each side.
1x – 3
43x=
3x 4(x – 3)=
3x 4x – 12=
3x – 4x – 12=
– x = – 12
x = 12
GUIDED PRACTICE for Example 4
7.
y – 3 7
y14=
SOLUTION
Write original proportion.
Cross Products Property
Multiply.
Subtract 7y from each side and add 42 to each side.
14y – 42 7y=
14y – 7y 42=
y = 6
y – 3 7
y14=
=14(y – 3) 7 y
Subtract , then divide
EXAMPLE 2 Use a ratio to find a dimension
SOLUTION
Painting
You are planning to paint a mural on a rectangular wall. You know that the perimeter of the wall is 484 feet and that the ratio of its length to its width is 9 : 2. Find the area of the wall.
Write expressions for the length and width. Because the ratio of length to width is 9 : 2, you can represent the length by 9x and the width by 2x.
STEP 1
EXAMPLE 2 Use a ratio to find a dimension
STEP 2
Solve an equation to find x.
Formula for perimeter of rectangle
Substitute for l, w, and P.Multiply and combine like terms.
Divide each side by 22.
=2l + 2w P=2(9x) + 2(2x) 484= 48422x
Evaluate the expressions for the length and width. Substitute the value of x into each expression.
STEP 3
The wall is 198 feet long and 44 feet wide, so its area is198 ft 44 ft = 8712 ft .2
= 22x
Length = 9x = 9(22) = 198Width = 2x = 2(22) = 44
EXAMPLE 3 Use extended ratios
Combine like terms.
SOLUTION
Triangle Sum Theorem
Divide each side by 6.= 30x=6x 180= 180
ox + 2x + 3x o oo
ALGEBRA The measures of the angles in CDE are in the extended ratio of 1 : 2 : 3. Find the measures of the angles.
Begin by sketching the triangle. Then use the extended ratio of 1 : 2 : 3 to label the measures as x° , 2x° , and 3x° .
The angle measures are 30 , 2(30 ) = 60 , and 3(30 ) = 90.o o o o o
ANSWER
GUIDED PRACTICE for Examples 2 and 3
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
Write expressions for the length and width. Because the ratio of length is 7 : 5, you can represent the length by 7x and the width by 5x.
STEP 1
GUIDED PRACTICE for Examples 2 and 3
STEP 2
Solve an equation to find x.
Formula for perimeter of rectangle
Substitute for l, w, and P.Multiply and combine like terms.
=2l + 2w P=2(7x) + 2(5x) 48= 4824x= 2x
Evaluate the expressions for the length and width. Substitute the value of x into each expression.
STEP 3
Length = 7x + 7(2) = 14 ftWidth = 5x + 5(2) = 10 ft
GUIDED PRACTICE for Examples 2 and 3
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
SOLUTION
Combine like terms.
Divide each side by 9.= 20x=9x 180=x + 3x + 5x 180o o oo
4. A triangle’s angle measures are in the extended ratio of 1 : 3 : 5. Find the measures of the angles.
x
3x 5x
The angle measures are 20 , 3(20 ) = 60 , and 5(20 ) = 100.o o o o o
ANSWER