Chapter 7

7-1 Ratios and Proportions

Course 3

Warm UpWarm Up

Problem of the DayProblem of the Day

Lesson PresentationLesson Presentation

Warm UpWrite each fraction in lowest terms.

Course 3


1416

1.

972

3.

2464

2.

45120

4.

78

38

18

38

Problem of the Day

A magazine has page numbers from 1 to 80. What fraction of those page numbers include the digit 5?

Course 3


1780

Learn to find equivalent ratios to create proportions.

Course 3


Vocabulary

ratioequivalent ratioproportion

Insert Lesson Title Here

Course 3


Course 3


Relative density is the ratio of the density of a substance to the density of water at 4°C. The relative density of silver is 10.5. This means that silver is 10.5 times as heavy as an equal volume of water.

The comparisons of water to silver in the table are ratios that are all equivalent.

42 g31.5 g21 g10.5 gSilver

4 g3 g2 g1 gWater

Comparisons of Mass of Equal Volumesof Water and Silver

Course 3


Ratios can be written in several ways. A colon is

often used. 90:3 and name the same ratio.

Reading Math

90 3

Course 3


A ratio is a comparison of two quantities by division. In one rectangle, the ratio of shaded squares to unshaded squares is 7:5. In the other rectangle, the ratio is 28:20. Both rectangles have equivalent shaded areas. Ratios that make the same comparison are equivalent ratios.

Course 3


Additional Example 1: Finding Equivalent Ratios

Find two ratios that are equivalent to each given ratio.

B.

1854

13

12848

83

A. =927

=9 • 227 • 2

=9 ÷ 927 ÷ 9

927

= Two ratios equivalent

to are and . 927

1854

13

Two ratios equivalent

to are and . 6424

12848

83

=64 • 224 • 2

=64 ÷ 824 ÷ 8

6424

=

6424

=

Multiply or divide the numerator and denominator by the same nonzero number.

Course 3


Try This: Example 1


B.

1632

24

6432

42

A. =816

=8 • 216 • 2

=8 ÷ 416 ÷ 4

816

= Two ratios equivalent

to are and . 816

1632

24

Two ratios equivalent

to are and . 3216

6432

42

=32 • 216 • 2

=32 ÷ 816 ÷ 8

3216

=

3216

=

Multiply or divide the numerator and denominator by the same nonzero number.

Course 3


Ratios that are equivalent are said to be proportional, or in proportion. Equivalent ratios are identical when they are written in simplest form.

Course 3


Additional Example 2: Determining Whether Two Ratios are in Proportion

Simplify to tell whether the ratios form a proportion.

1215

B. and 2736

327

A. and 218

Since ,

the ratios are in

proportion.

19

= 19

19

=3 ÷ 327 ÷ 3

327

=

19

=2 ÷ 218 ÷ 2

218

=

45=

12 ÷ 315 ÷ 3

1215

=

34=

27 ÷ 936 ÷ 9

2736

=

Since ,

the ratios are not

in proportion.

45 3

4

Course 3


Try This: Example 2


1449

B. and 1636

Since ,

the ratios are in

proportion.

15

= 15

15

=3 ÷ 315 ÷ 3

315

=

15

=9 ÷ 945 ÷ 9

945

=

27

=14 ÷ 749 ÷ 7

1449

=

49=

16 ÷ 436 ÷ 4

1636

=

Since ,

the ratios are not

in proportion.

27 4

9

315

A. and 945

Course 3


Additional Example 3: Earth Science Application

At 4°C, four cubic feet of silver has the same mass as 42 cubic feet of water. At 4°C, would 210 cubic feet of water have the same mass as 20 cubic feet of silver?

4 ÷ 242 ÷ 2

?= 20 ÷ 10210 ÷ 10

221

= 221

442

?= 20210

Since ,

210 cubic feet of water would have the same mass at 4°C as 20 cubic feet of silver.

221

= 221

Course 3


Try This: Example 3

At 4°C, two cubic feet of silver has the same mass as 21 cubic feet of water. At 4°C, would 105 cubic feet of water have the same mass as 10 cubic feet of silver?

?= 10 ÷ 5105 ÷ 5

221

221

= 221

221

?= 10105

Since ,

105 cubic feet of water would have the same mass at 4°C as 10 cubic feet of silver.

221

= 221

Lesson Quiz: Part 1


Course 3


85

85

= ; yes


415

1.

821

2.

1610

3.

3624

4.


830

1245

Possible answer: ,

1642

2463

Possible answer: ,

and 32 20

and 28 18

32

149

; no

Lesson Quiz: Part 2


Course 3


5. Kate poured 8 oz of juice from a 64 oz bottle. Brian poured 16 oz of juice from a 128 oz bottle. What ratio of juice is missing from each bottle? Are the ratios proportional?

864

16128

and ; yes, both equal 1 8

7-2 Ratios, Rates, and Unit Rates

Course 3

Warm UpWarm Up



Warm UpDivide. Round answers to the nearest tenth.

1. 2.

3. 4.

23.3 3.5

23.8

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23.9

420 18

73 21

380 16

430 18

Problem of the Day

There are 3 bags of flour for every 2 bags of sugar in a freight truck. A bag of flour weighs 60 pounds, and a bag of sugar weighs 80 pounds. Which part of the truck’s cargo is heavier, the flour or the sugar?flour

Course 3


Learn to work with rates and ratios.

Course 3


Vocabulary

rateunit rateunit price


Course 3


Course 3


An aspect ratio describes a screen by comparing its width to its height.

Movie and television screens range in shape from almost perfect squares to wide rectangles.

Common aspect ratios are 4:3, 37:20, 16:9, and 47:20.

Additional Example 1A: Ordering Ratios

A. Order the ratios 4:3, 23:10, 13:9, and 47:20 from the least to greatest.


Course 3


4:3 =

23:10 =

13:9 =

47:20 =

4 3

4 3

Divide. = 1.3 1

= 1.3

23 10

13 947 20

= 1.4

= 2.3

= 2.35

The ratios in order from least to greatest are 4:3, 13:9, 23:10, and 47:20.

The decimals in order are 1.3, 1.4, 2.3, and 2.35.

B. A television has screen width 20 in. and height 15 in. What is the aspect ratio of this screen?


Course 3


Additional Example 1B: Ordering Ratios

The ratio of the width to the height is 20:15.

The screen has the aspect ratio 4:3.

The ratio can be simplified: 20 15

5(4) 5(3)

4 3

20 15

= = .

A. Order the ratios 2:3, 35:14, 5:3, and 49:20 from the least to greatest.


Try This: Example 1A

Course 3


2:3 =

35:14 =

5:3 =

49:20 =

2 3

2 3

Divide. = 0.6 1

= 0.6

35 14

5

3 49 20

= 1.6

= 2.5

= 2.45

The ratios in order from least to greatest are 2:3, 5:3, 49:20, 35:14.

The decimals in order are 0.6, 1.6, 2.45, and 2.5.

B. A movie theater has a screen width 36 ft. and height 20ft. What is the aspect ratio of this screen?


Try This: Example 1B

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The ratio of the width to the height is 36:20.

The screen has the aspect ratio 9:5.

The ratio can be simplified: 36 20

4(9) 4(5)

9 5

36 20

= = .

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Ratio: 903

Rate: 90 miles3 hours

Read as “90 miles per 3

hours.”

A rate is a comparison of two quantities that have different units.

A ratio is a comparison of two quantities.

Course 3


Unit rates are rates in which the second quantity is 1.

unit rate: 30 miles,1 hour

or 30 mi/h

The ratio 903

can be simplified by dividing:

903

= 301


Additional Example 2: Using a Bar Graph to Determine Rates

Use the bar graph to find the number of acres, to the nearest acre, destroyed in Nevada and Alaska per week.

Nevada =

12,308 acres1 week

Course 3


640,000 acres52 weeks


Additional Example 2 Continued

Use the bar graph to find the number of acres, to the nearest acre, destroyed in Nevada and Alaska per week.

Alaska =

14,423 acres1 week

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Montana =

18,269 acres1 week

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Try This: Example 2

Use the bar graph to find the number of acres, to the nearest acre, destroyed in Montana and Idaho per week.


Idaho =

26,923 acres1 week

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1,400,000 acres52 weeks

Try This: Example 2 Continued

Use the bar graph to find the number of acres, to the nearest acre, destroyed in Montana and Idaho per week.

Unit price is a unit rate used to compare costs per item.

Course 3


Pens can be purchased in a 5-pack for $1.95 or a 15-pack for $6.20. Which is the better buy?

Additional Example 3A: Finding Unit Prices to Compare Costs

Divide the price by the number of pens.

price for packagenumber of pens

$1.955

= $0.39

price for packagenumber of pens

= $6.2015

$0.41

Course 3


The better buy is the 5-pack for $1.95.

Jamie can buy a 15-oz jar peanut butter for $2.19 or a 20-oz jar for $2.78. Which is the better buy?

Additional Example 3B: Finding Unit Prices to Compare Costs

$2.1915

= $0.15

= $2.7820

$0.14

Course 3


The better buy is the 20-oz jar for $2.78.

price for jarnumber of ounces

price for jarnumber of ounces

Divide the price by the number of ounces.

Golf balls can be purchased in a 3-pack for $4.95 or a 12-pack for $18.95. Which is the better buy?


Course 3


Divide the price by the number of balls.

price for packagenumber of balls

$4.953

= $1.65

price for packagenumber of balls

= $18.9512

$1.58

The better buy is the 12-pack for $18.95.


John can buy a 24 oz bottle of ketchup for $2.19 or a 36 oz bottle for $3.79. Which is the better buy?

Course 3


$2.1924

= $0.09

= $3.7936

$0.11

The better buy is the 24-oz jar for $2.19.

price for bottlenumber of ounces

price for bottlesnumber of ounces

Divide the price by the number of ounces.

Lesson Quiz1. At a family golf outing, a father drove the ball

285 ft. His daughter drove the ball 95 ft. Express the ratio of the father’s distance to his daughter’s in simplest terms.

2. Find the unit price of 6 stamps for $2.22.

3. Find the unit rate of 8 heartbeats in 6 seconds.

4. What is the better buy, a half dozen carnations

for $4.75 or a dozen for $9.24?

5. Which is the better buy, four pens for $5.16 or

a ten-pack for $12.90?

$0.37 per stamp

3:1


1.3 beats/s

a dozen

Course 3


They cost the same.

7-3 Analyze Units

Course 3

Warm UpWarm Up



Warm UpFind each unit rate.

1. Jump rope 192 times in 6 minutes

2. Four pounds of bananas for $2.36

3. 16 anchor bolts for $18.56

4. 288 movies on 9 shelves

32 jumps/min

$0.59/lb

$1.16/bolt

Course 3

7-3 Analyze Units

32 movies/shelf

Problem of the Day

Replace each • with a digit from 0 to 6 to make equivalent ratios. Use each digit only once.

Course 3

7-3 Analyze Units

••••

•••

= Possible answer:1365

420

=

Learn to use one or more conversion factors to solve rate problems.

Course 3

7-3 Analyze Units

Vocabulary

conversion factor


Course 3

7-3 Analyze Units

Course 3

7-3 Analyze Units

You can measure the speed of an object by using a strobe lamp and a camera in a dark room. Each time the lamp flashes, the camera records the object’s position.

Problems often require dimensional analysis, also called unit analysis, to convert from one unit to another unit.

Course 3

7-3 Analyze Units

To convert units, multiply by one or more ratios of equal quantities called conversion factors.

For example, to convert inches to feet you would use the ratio below as a conversion factor.

1 ft12 in.

Course 3

7-3 Analyze Units

Multiplying by a conversion factor is like multiplying by a fraction that reduces to 1, such as .5

5

12 in.12 in.

1 ft1 ft

= , or = 11 ft12 in.

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7-3 Analyze Units

The conversion factor

• must introduce the unit desired in the answer and

• must cancel the original unit so that the unit desired is all that remains.

Helpful Hint

Find the appropriate factor for each conversion.

Additional Example 1: Finding Conversion Factors

Course 3

7-3 Analyze Units

A. feet to yards

B. pounds to ounces

1 yd3 ft

There are 3 feet in 1 yard. To convert feet to

yards, multiply the number of feet by .

16 oz1 lb

There are 16 ounces in 1 pound. To convert

pounds to ounces, multiply the number of

pounds by .

Try This: Example 1



Course 3

7-3 Analyze Units

A. minutes to seconds

B. hours to days

60 sec1 min

There are 60 seconds in 1 minute. To convert

minutes to seconds, multiply the number of

minutes by .

1 day 24 h

There are 24 hours in 1 day. To convert

hours to days, multiply the number hours by

.

The average American uses 580 pounds of paper per year. Find the number of pounds of paper the average American uses per month, to the nearest tenth.

Additional Example 2: Using Conversion Factors to Solve Problems

Course 3

7-3 Analyze Units

The problem gives the ratio 580 pounds to 1 year and asks for an answer in pounds per month.

580 lb 1 yr

1 yr 12 mo

580 lb 12 mo

=

= 48.3 lb per month

Multiply the ratio by the conversion factor

Cancel yr units.

Divide 580 by 12.

The average American uses 580 pounds of paper per year. Find the number of pounds of paper the average American uses per month, to the nearest tenth.


Course 3

7-3 Analyze Units

The average American uses 48.3 pounds of paper per month.

Try This: Example 2

Sam drives his car 23,000 miles per year. Find the number of miles he drives per month.


Course 3

7-3 Analyze Units

The problem gives the ratio 23,000 miles to 1 year and asks for an answer in miles per month.

23,000 mi 1 yr

1 yr 12 mo

23,000 mi 12 mo

=

= 1916.6 per month

Multiply the ratio by the conversion factor

Cancel yr units.

Divide 23,000 by 12.

Sam drives his car about 1917 miles per month.

11 Understand the Problem

Course 3

7-3 Analyze Units

Additional Example 3: Problem Solving Application

A car traveled 60 miles on a road in 2 hours. How many feet per second was the car traveling?

The problem is stated in units of miles and hours. The question asks for the answer in units of feet and seconds. You will need to use several conversion factors.List the important information:

• Miles to feet5280 ft

1 mi• Hours to minutes

• Minutes to seconds 1 min60 s

1 h60 min

Multiply by each conversion factor separately, or simplify the problem and multiply by several conversion factors at once.

Course 3

7-3 Analyze Units

22 Make a Plan


First, convert 60 miles in 2 hours into a unit rate.

Course 3

7-3 Analyze Units

Solve33

60 mi2 h

= (60÷2) mi(2÷2) h

= 30 mi1 h

Create a single conversion factor to convert hours directly to seconds:

hours to seconds = • 1 min60 s

Set up the conversion factors.

minutes to seconds 1 min60 s

hours to minutes 1 h60 min

1 h60 min

1 h3600 s

=

30 mi1 h

• 5280 ft1 mi

• 1 h 3600 s

Course 3

7-3 Analyze Units

Solve33

Do not include the numbers yet.

Notice what happens to the units.

30 • 5280 ft • 1 1 • 1 • 3600 s

= 158,400 ft3600 s

= 44 ft1 s

The car was traveling 44 feet per second.

Simplify. Only remains.fts

Multiply.

mih

•ftmi

• hs

• •30 mi1 h

5280 ft1 mi

1 h 3600 s

A rate of 44 ft/s is less than 50 ft/s. A rate of 60 miles in 2 hours is 30 mi/h or 0.5 mi/min.

Course 3

7-3 Analyze Units

44 Look Back

Since 0.5 mi/min is less than 3000 ft/60 s or 50 ft/s and 44 ft/s is less than 50 ft/s, then 44 ft/s is a reasonable answer.

11 Understand the Problem

Course 3

7-3 Analyze UnitsTry This: Example 3

A train traveled 180 miles on a railroad track in 4 hours. How many feet per second was the train traveling?

The problem is stated in units of miles and hours. The question asks for the answer in units of feet and seconds. You will need to use several conversion factors.List the important information:

• Miles to feet5280 ft

1 mi• Hours to minutes

• Minutes to seconds 1 min60 s

1 h60 min

Multiply by each conversion factor separately, or simplify the problem and multiply by several conversion factors at once.

Course 3

7-3 Analyze Units

22 Make a Plan


First, convert 180 miles in 4 hours into a unit rate.

Course 3

7-3 Analyze Units

Solve33

180 mi4 h

= (180 ÷ 4) mi(4 ÷ 4) h

= 45 mi1 h

Create a single conversion factor to convert hours directly to seconds:

hours to seconds = • 1 min60 s

Set up the conversion factors.

minutes to seconds 1 min60 s

hours to minutes 1 h60 min

1 h60 min

1 h3600 s

=

45 mi1 h

• 5280 ft1 mi

• 1 h 3600 s

Course 3

7-3 Analyze Units

Solve33

Do not include the numbers yet.

Notice what happens to the units.

45 • 5280 ft • 1 1 • 1 • 3600 s

= 237,600 ft3600 s

= 66 ft1 s

The train was traveling 66 feet per second.

Simplify. Only remains.fts

Multiply.

mih

•ftmi

• hs

• •45 mi1 h

5280 ft1 mi

1 h 3600 s

A rate of 66 ft/s is more than 50 ft/s. A rate of 180 miles in 4 hours is 45 mi/h or 0.75 mi/min.

Course 3

7-3 Analyze Units

44 Look Back

Since 0.75 mi/min is more than 3000 ft/60 s or 50 ft/s and 66 ft/s is more than 50 ft/s, then 66 ft/s is a reasonable answer.

Additional Example 4: Physical Science Application

Course 3

7-3 Analyze Units

A strobe lamp can be used to measure the speed of an object. The lamp flashes every of a second. A camera records the object moving 52 cm between flashes. How fast is the object moving in m/s?

1 100

distance .time

Use rate =52 cm1

100s

It may help to eliminate the fraction first.


Course 3

7-3 Analyze Units

1 100

Multiply top and bottom by 100.

5200 cm1 s

=

52 cm1

100s

= 100 • 52 cm 1

100 s 100 •

Now convert centimeters to meters.


Course 3

7-3 Analyze Units

5200 cm1 s

Multiply by the conversion factor.

5200 m100 s

=52 m1 s

=

The object is traveling 52 m/s.

5200 cm1 s

= • 1 m100 cm

Try This: Example 4

Course 3

7-3 Analyze Units

A strobe lamp can be used to measure the speed of an object. The lamp flashes every of a second. A camera records the object moving 65 cm between flashes. How fast is the object moving in m/s?

1 100

distance .time

Use rate =65 cm1

100s

It may help to eliminate the fraction first.


Course 3

7-3 Analyze Units

1 100

Multiply top and bottom by 100.

6500 cm1 s

=

65 cm1

100s

= 100 • 65 cm 1

100 s 100 •

Now convert centimeters to meters.


Course 3

7-3 Analyze Units

6500 cm1 s

Multiply by the conversion factor.

6500 m100 s

=65 m1 s

=

The object is traveling 65 m/s.

6500 cm1 s

= • 1 m100 cm

The rate 1 knot equals 1 nautical mile per hour. One nautical mile is 1852 meters. What is the speed in kilometers per hour of a ship traveling at 5 knots?

Additional Example 5: Transportation Application

Course 3

7-3 Analyze Units

5 knots = 5 nautical mi/h

Set up the units to obtain in your answer.kmh

Examine the units.

5 • 1852 • 1 km1 h • 1 • 1000

= = 9260 km1000 h

=9.26 km1 h

The ship is traveling 9.26 km/h.

nautical mih

• mnautical mi

• kmm

5 nautical mi1 h

• •1852 m1 nautical mi

1 km1000 m

Course 3

7-3 Analyze Units

Try This: Example 5

The rate 1 knot equals 1 nautical mile per hour. One nautical mile is 1852 meters. What is the speed in kilometers per hour of a ship traveling at 9 knots?9 knots = 9 nautical mi/h

Set up the units to obtain in your answer.kmh

Examine the units.

9 • 1852 • 1 km1 h • 1 • 1000

= =16,668 km1000 h

16.67 km1 h

The ship is traveling about 16.67 km/h.

nautical mih

• mnautical mi

• kmm

9 nautical mi1 h

• •1852 m1 nautical mi

1 km1000 m

Lesson Quiz


1. kilograms to grams

2. pints to gallons

3. You drive 136 miles from your house to your aunt’s

house at the lake. You use 8 gallons of gas. How many

yards does your car get to the gallon?

4. A cheetah was timed running 200 yards in 6

seconds. What was the average speed in miles per

hour?


Course 3

7-3 Analyze Units

1000 gkg

1 gal8 pt

29,920 ydgal

≈ 68 mi/h

7-4 Solving Proportions

Course 3

Warm UpWarm Up



Warm UpFind two ratios that are equivalent to each given ratio.

Course 3


35

1.

4530

3. 9060

32

,

1012

2. 2024

56

,

89

4. 2427

1618

,

915

610

,Possible answers:

Problem of the Day

Replace each • with a digit from 1 to 7 to write a proportion. Use each digit once. The digits 2 and 3 are already shown.

Course 3


••

•23

••= 14

723

56=Possible answer:

Learn to solve proportions.

Course 3


Vocabulary

cross product


Course 3


Course 3


Unequal masses will not balance on a fulcrum if they are an equal distance from it; one side will go up and the other side will go down.

Unequal masses will balance when the following proportions is true:

mass 2length 1

mass 1length 2

=

Mass 1

Mass 2

Fulcrum

Length 1 Length 2

Course 3


One way to find whether ratios, such as those on the previous slide, are equal is to find a common denominator. The ratios are equal if their numerators are equal after the fractions have been rewritten with a common denominator.

7296

68

= 7296

912

=

912

68

=

Course 3


Course 3


The cross product represents the numerator of the fraction when a common denominator is found by multiplying the denominators.

Helpful Hint

Course 3


Tell whether the ratios are proportional.

410

615

A.

Since the cross products are equal, the ratios are proportional.

60

=?

Additional Example 1A: Using Cross Products to Identify Proportions

60 = 60

Find cross products.60410

615

Course 3


A mixture of fuel for a certain small engine should be 4 parts gasoline to 1 part oil. If you combine 5 quarts of oil with 15 quarts of gasoline, will the mixture be correct?

4 parts gasoline1 part oil

=? 15 quarts gasoline5 quarts oil

4 • 5 = 20 1 • 15 = 15

20 ≠ 15

The ratios are not equal. The mixture will not be correct.

Set up ratios.

Find the cross products.

Additional Example 1B: Using Cross Products to Identify Proportions

Course 3


Course 3


Tell whether the ratios are proportional.


Since the cross products are equal, the ratios are proportional.

20

20 = 20

Find cross products.2024

510

24

510

A. =?

Course 3


A mixture for a certain brand of tea should be 3 parts tea to 1 part sugar. If you combine 4 tablespoons of sugar with 12 tablespoons of tea, will the mixture be correct?


3 parts tea 1 part sugar

=? 12 tablespoons tea4 tablespoons sugar

3 • 4 = 12 1 • 12 = 12

12 = 12

The ratios are equal. The mixture will be correct.

Set up ratios.


Course 3


When you do not know one of the four numbers in a proportion, set the cross products equal to each other and solve.

Course 3


Solve the proportion.

6p = 12 • 5

p = 10

6p = 60


Solve.

Additional Example 2: Solving Proportions

56

p12

=

; the proportion checks.56

1012

=

Course 3



14 • 3 = 2g

21 = g

42 = 2g


Solve.

Try This: Example 2

23

14g

=

; the proportion checks.23

1421

=

Course 3


Allyson weighs 55 lbs and sits on a seesaw 4 ft away from its center. If Marco sits 5 ft away from the center and the seesaw is balanced, how much does Marco weigh?

5x5

2205

=

44 = x

Set up the proportion.

Let x represent Marco’s weight.


Multiply.

Solve. Divide both sides by 5.

Marco weighs 44 lb.

Additional Example 3: Physical Science Application

220 = 5x

55 • 4 = 5x

x4

555

=

mass 1length 2

= mass 2length 1

Course 3


Robert weighs 90 lbs and sits on a seesaw 5 ft away from its center. If Sharon sits 6 ft away from the center and the seesaw is balanced, how much does Sharon weigh?

Try This: Example 3

6x6

4506

=

75 = x

Set up the proportion.

Let x represent Sharon’s weight.


Multiply.

Solve. Divide both sides by 6.

Sharon weighs 75 lb.

450 = 6x

90 • 5 = 6x

x5

906

=

mass 1length 2

= mass 2length 1

Lesson Quiz


Course 3


Tell whether each pair of ratios is proportional.

4842 =? 16

141. 20

15 =? 34

2.

Solve each proportion.

3. 4.

5. Two weights are balanced on a fulcrum. If a 6lb weight is positioned 1.5 ft from the fulcrum, at what distance from the fulcrum must an 18 lb weight be placed to keep the weights balanced?

yes no

n = 30 n = 16

0.5 ft

4518

n12 = n

2469 =

7-5 Dilations

Course 3

Warm UpWarm Up



Warm UpMultiply.

1. 4 2. 12

3. 24 4. –36

Course 3

7-5 Dilations

3

18

10

9

–27

30

3 4

3 4

3 4

3 4

5. 4 2.5 6. 12 2.5

Problem of the Day

Every day, a plant grows to three times its size. Every night, it shrinks to half its size. After three days and nights, it is 6.75 in. tall. How tall was the plant at the start?

2 in.

Course 3

7-5 Dilations

Learn to identify and create dilations of plane figures.

Course 3

7-5 Dilations

Vocabulary

dilationscale factorcenter of dilation


Course 3

7-5 Dilations

Course 3

7-5 Dilations

Your pupils are the black areas in the center of your eyes. When you go to the eye doctor, the doctor may dilate your pupils, which makes them larger.

Course 3

7-5 Dilations

Translations, reflections, and rotations are transformations that do not change the size or shape of a figure. A dilation is a transformation that changes the size, but not the shape, of a figure. A dilation can enlarge or reduce a figure.

Course 3

7-5 Dilations

A scale factor describes how much a figure is enlarged or reduced. A scale factor can be expressed as a decimal, fraction, or percent. A 10% increase is a scale factor of 1.1, and a 10% decrease is a scale factor of 0.9.


Course 3

7-5 Dilations

A scale factor between 0 and 1 reduces a figure. A scale factor greater than 1 enlarges it.

Helpful Hint

Tell whether each transformation is a dilation.


Course 3

7-5 Dilations

The transformation is a dilation.

The transformation is not a dilation. The figure is distorted.

Additional Example 1A & 1B: Identifying Dilations

A. B.



Course 3

7-5 Dilations



Additional Example 1C & 1D: Identifying Dilations

C. D.

Course 3

7-5 Dilations


A'

B' C'

A

B C

A.

B

A

C

A'

B' C'



Try This: Example 1A & 1B

B.

Course 3

7-5 Dilations


C.



Try This: Example 1C & 1D

A'

B' C'

A

B C

D.

A'

B' C'

A

B C


Course 3

7-5 Dilations

Every dilation has a fixed point that is the center of dilation. To find the center of dilation, draw a line that connects each pair of corresponding vertices. The lines intersect at one point. This point is the center of dilation.

Dilate the figure by a scale factor of 1.5 with P as the center of dilation.


Course 3

7-5 Dilations

Additional Example 2: Dilating a Figure

Multiply each side by 1.5.


Course 3

7-5 Dilations

Dilate the figure by a scale factor of 0.5 with G as the center of dilation.

G

F H

2 cm 2 cm

2 cm

Multiply each side by 0.5.

Try This: Example 2

G

F H

2 cm 2 cm

2 cm

F’ H’1 cm

1 cm

1 cm


Course 3

7-5 DilationsAdditional Example 3A: Using the Origin as the Center of

DilationDilate the figure in Example 3A on page 363 by a scale factor of 2. What are the vertices of the image?

Multiply the coordinates by 2 to find the vertices of the image.

A(4, 8) A’(4 2, 8 2) A’(8, 16)

B(3, 2) B’(3 2, 2 2) B’(6, 4)

C(5, 2) C’(5 2, 2 2) C’(10, 4)

The vertices of the image are A’(8, 16), B’(6, 4), and C’(10, 4).

ABC A’B’C’


Course 3

7-5 Dilations

Additional Example 3B: Using the Origin as the Center of Dilation

Dilate the figure in Example 3B by a scale factor

of . What are the vertices of the image?1 3


ABC A’B’C’

A(3, 9) A’(3 , 9 ) A’(1, 3) 1 3

1 3

B(9, 6) B’(9 , 6 ) B’(3, 2) 1 3

1 3

C(6, 3) C’(6 , 3 ) C’(2, 1) 1 3

1 3

Multiply the coordinates by to find the vertices of the image.

1 3


Course 3

7-5 Dilations


Dilate the figure by a scale factor of 2. What are the vertices of the image?

2

4

2 4 6 8 100

6

8

10

B

C

A


Course 3

7-5 Dilations

Try This: Example 3A Continued

A(2, 2) A’(2 2, 2 2) A’(4, 4)

B(4, 2) B’(4 2, 2 2) B’(8, 4)

C(2, 4) C’(2 2, 4 2) C’(4, 8)

ABC A’B’C’



Course 3

7-5 Dilations


2

4

2 4 6 8 100

6

8

10

B’

C’

A’

B

C

A


Course 3

7-5 Dilations


Dilate the figure by a scale factor of 0.5. What are the vertices of the image?

2

4

2 4 6 8 100

6

8

10

B

C

A


Course 3

7-5 Dilations

Try This: Example 3B Continued

A(4, 5) A’(4 0.5, 5 0.5) A’(2, 2.5)

B(8, 5) B’(8 0.5, 5 0.5) B’(4, 2.5)

C(4, 9) C’(4 0.5, 9 0.5) C’(2, 4.5)

ABC A’B’C’

The vertices of the image are A’(2, 2.5), B’(4, 2.5), and C’(2, 4.5).


Course 3

7-5 Dilations

Try This: Example 3B Continued

2

4

2 4 6 8 100

6

8

10

B

C

A

B’

C’

A’

Lesson Quiz


Course 3

7-5 Dilations

1. Tell whether the transformation is a dilation. A(0, 4) B(5,5) C(3,3) A’(0, 8) B’(10, 10) C’(6, 6)

2 4 6

-2

-4

-6

P

AB

CC’

B’ A’

2. Dilate the figure by a scale factor of 1.5 with P as the center of dilation.

3. Dilate the figure by a scale factor of 2 with the origin as the center of dilation. What are the coordinates of the image? A(2,4) B(5,6) C(6,1) A’(4,8) B’(10,12) C’(12,2)

yes

7-6 Similar Figures

Course 3

Warm UpWarm Up



Warm UpSolve each proportion.

Course 3

7-6 Similar Figures

b30

39

=1. 5635

y5

=2.

412

p9

=3. 56m

2826

=4.

b = 10 y = 8

p = 3 m = 52

Problem of the Day

A plane figure is dilated and gets 50% larger. What scale factor should you use to dilate the figure back to its original size? (Hint: The answer is not ).

Course 3

7-6 Similar Figures

12

23

Learn to determine whether figures are similar, to use scale factors, and to find missing dimensions in similar figures.

Course 3

7-6 Similar Figures

Vocabulary

similar


Course 3

7-6 Similar Figures

Course 3

7-6 Similar Figures

The heights of letters in newspapers and on billboards are measured using points and picas. There are 12 points in 1 pica and 6 picas in one inch.

A letter 36 inches tall on a billboard would be 216 picas, or 2592 points. The first letter in this paragraph is 12 points.

Course 3

7-6 Similar Figures

Congruent figures have the same size and shape. Similar figures have the same shape, but not necessarily the same size. The A’s in the table are similar. The have the same shape, but they are not the same size.

The ratio formed by the corresponding sides is the scale factor.

Course 3

7-6 Similar Figures

Additional Example 1: Using Scale Factors to Find Missing Dimensions

A picture 10 in. tall and 14 in. wide is to be scaled to 1.5 in. tall to be displayed on a Web page. How wide should the picture be on the Web page for the two pictures to be similar?To find the scale factor, divide the known measurement of the scaled picture by the corresponding measurement of the original picture.

0.15 0.151.510

=

Then multiply the width of the original picture by the scale factor.2.1 14 • 0.15 = 2.1

The picture should be 2.1 in. wide.

Course 3

7-6 Similar Figures

Try This: Example 1

A painting 40 in. tall and 56 in. wide is to be scaled to 10 in. tall to be displayed on a poster. How wide should the painting be on the poster for the two pictures to be similar?To find the scale factor, divide the known measurement of the scaled painting by the corresponding measurement of the original painting.0.25 0.2510

40=

Then multiply the width of the original painting by the scale factor.14 56 • 0.25 = 14

The painting should be 14 in. wide.

Course 3

7-6 Similar Figures

Additional Example 2: Using Equivalent Ratios to Find Missing Dimensions

A T-shirt design includes an isosceles triangle with side lengths 4.5 in, 4.5 in., and 6 in. An advertisement shows an enlarged version of the triangle with two sides that are each 3 ft. long. What is the length of the third side of the triangle in the advertisement?

Set up a proportion.

6 in.x ft

4.5 in.3 ft

=

4.5 in. • x ft = 3 ft • 6 in. Find the cross products.

4.5 in. • x ft = 3 ft • 6 in. in. • ft is on both sides

Course 3

7-6 Similar Figures

4.5x = 3 • 6

4.5x = 18

x = = 4184.5

Cancel the units.

Multiply

Solve for x.


The third side of the triangle is 4 ft long.

Course 3

7-6 Similar Figures

Try This: Example 2

Set up a proportion.

24 ftx in.

18 ft4 in.

=

18 ft • x in. = 24 ft • 4 in. Find the cross products.

18 ft • x in. = 24 ft • 4 in. in • ft is on both sides

A flag in the shape of an isosceles triangle with side lengths 18 ft, 18 ft, and 24 ft is hanging on a pole outside a campground. A camp t-shirt shows a smaller version of the triangle with two sides that are each 4 in. long. What is the length of the third side of the triangle on the t-shirt?

Course 3

7-6 Similar Figures

18x = 24 • 4

18x = 96

x = 5.39618

Cancel the units.

Multiply

Solve for x.


The third side of the triangle is about 5.3 in. long.

Course 3

7-6 Similar Figures

The following are matching, or corresponding:A and X

Remember!

AB and XY

BC and YZ

AC and XZ

A

C BZ Y

X

C and Z

B and Y

Course 3

7-6 Similar Figures

Additional Example 3: Identifying Similar Figures

Which rectangles are similar?

Since the three figures are all rectangles, all the angles are right angles. So the corresponding angles are congruent.

Course 3

7-6 Similar Figures


50 48

The ratios are equal. Rectangle J is similar to rectangle K. The notation J ~ K shows similarity.

The ratios are not equal. Rectangle J is not similar to rectangle L. Therefore, rectangle K is not similar to rectangle L.

20 = 20

length of rectangle Jlength of rectangle K

width of rectangle Jwidth of rectangle K

10 5

42

? =

length of rectangle Jlength of rectangle L

width of rectangle Jwidth of rectangle L

1012

45

?=

Compare the ratios of corresponding sides to see if they are equal.

Course 3

7-6 Similar Figures

Try This: Example 3

Which rectangles are similar?

A8 ft

4 ft

B6 ft

3 ft

C5 ft

2 ft

Since the three figures are all rectangles, all the angles are right angles. So the corresponding angles are congruent.

Course 3

7-6 Similar Figures

Try This: Example 3

16 20

The ratios are equal. Rectangle A is similar to rectangle B. The notation A ~ B shows similarity.

The ratios are not equal. Rectangle A is not similar to rectangle C. Therefore, rectangle B is not similar to rectangle C.

24 = 24

length of rectangle Alength of rectangle B

width of rectangle Awidth of rectangle B

8 6

43

? =

length of rectangle Alength of rectangle C

width of rectangle Awidth of rectangle C

8 5

42

?=

Compare the ratios of corresponding sides to see if they are equal.

Course 3

7-6 Similar Figures

Lesson QuizUse the properties of similar figures to answer each question.

1. A rectangular house is 32 ft wide and 68 ft long. On a blueprint, the width is 8 in. Find the length on the blueprint.

2. Karen enlarged a 3 in. wide by 5 in. tall photo into a poster. If the poster is 2.25 ft wide, how tall is it?

3. Which rectangles are similar?

17 in.

3.75 ft

A and B are similar.

7-7 Scale Drawings

Course 3

Warm UpWarm Up



Warm UpEvaluate the following for x = 16.

1. 3x 2. x

Evaluate the following for x = .

3. 10x 4. x

48 12

4

Course 3

7-7 Scale Drawings

34

25

14

110

Problem of the Day

An isosceles triangle with a base length of 6 cm and side lengths of 5 cm is dilated by a scale factor of 3. What is the area of the image?

108 cm2

Course 3

7-7 Scale Drawings

Learn to make comparisons between and find dimensions of scale drawings and actual objects.

Course 3

7-7 Scale Drawings

Vocabulary

scale drawingscalereductionenlargement


Course 3

7-7 Scale Drawings

Course 3

7-7 Scale Drawings

A scale drawing is a two-dimensional drawing that accurately represents an object. The scale drawing is mathematically similar to the object.

A scale gives the ratio of the dimensions in the drawing to the dimensions of the object. All dimensions are reduced or enlarged using the same scale. Scales can use the same units or different units.


The scale a:b is read “a to b.” For example, the scale 1 cm:3 ft is read “one centimeter to three feet.”

Reading Math

Course 3

7-7 Scale Drawings

Scale Interpretation

1:20 1 unit on the drawing is 20 units.

1 cm: 1 m 1 cm on the drawing is 1 m.

in. = 1 ft in. on the drawing is 1 ft.1 4

1 4

A. The length of an object on a scale drawing is 2 cm, and its actual length is 8 m. The scale is 1 cm: __ m. What is the scale?

Additional Example 1A: Using Proportions to Find Unknown Scales or Lengths

Course 3

7-7 Scale Drawings

1 cmx m = 2 cm

8 m Set up proportion using scale length .actual length

1 8 = x 2 Find the cross products.

8 = 2x


The scale is 1 cm:4 m.

4 = x

B. The length of an object on a scale drawing is 1.5 inches. The scale is 1 in:6 ft. What is the actual length of the object?

Additional Example 1B: Using Proportions to Find Unknown Scales or Lengths

Course 3

7-7 Scale Drawings

1 in.6 ft = 1.5 in.

x ft Set up proportion using scale length .actual length

1 x = 6 1.5 Find the cross products.

x = 9 Solve the proportion.

The actual length is 9 ft.

A. The length of an object on a scale drawing is 4 cm, and its actual length is 12 m. The scale is 1 cm: __ m. What is the scale?


Course 3

7-7 Scale Drawings

1 cmx m = 4 cm

12 m Set up proportion using scale length .actual length

1 12 = x 4 Find the cross products.

12 = 4x


The scale is 1 cm:3 m.

3 = x

B. The length of an object on a scale drawing is 2 inches. The scale is 1 in:4 ft. What is the actual length of the object?


Course 3

7-7 Scale Drawings

1 in.4 ft = 2 in.

x ft Set up proportion using scale length .actual length

1 x = 4 2 Find the cross products.

x = 8 Solve the proportion.

The actual length is 8 ft.


Course 3

7-7 Scale Drawings

A scale drawing that is smaller than the actual object is called a reduction. A scale drawing can also be larger than the object. In this case, the drawing is referred to as an enlargement.

Under a 1000:1 microscope view, an amoeba appears to have a length of 8 mm. What is its actual length?

Additional Example 2: Life Sciences Application

Course 3

7-7 Scale Drawings

scale length actual length

1000 x = 1 8 Find the cross products.

x = 0.008

The actual length of the amoeba is 0.008 mm.

1000 1 = 8 mm

x mm


Under a 10,000:1 microscope view, a fiber appears to have length of 1mm. What is its actual length?

Try This: Example 2

Course 3

7-7 Scale Drawings


10,000 x = 1 1 Find the cross products.

x = 0.0001

The actual length of the fiber is 0.0001 mm.

10,000 1 = 1 mm

x mm



Course 3

7-7 Scale Drawings

A drawing that uses the scale in. = 1 ft

is said to be in in. scale. Similarly, a

drawing that uses the scale in. = 1 ft is

in in. scale.

1 4

1 4

1 2

1 2

Additional Example 3A: Using Scales and Scale Drawings to Find Heights

Course 3

7-7 Scale Drawings


0.25 x = 1 4 Find the cross products.

x = 16

The wall is 16 ft tall.

0.25 in. 1 ft = 4 in.

x ft.

A. If a wall in a in. scale drawing is 4 in. tall, how tall is the actual wall?

14

Length ratios are equal.


B. How tall is the wall if a in. scale is used?

Additional Example 3B: Using Scales and Scale Drawings to Find Heights

Course 3

7-7 Scale Drawings

12


0.5 x = 1 4 Find the cross products.

x = 8

The wall is 8 ft tall.

0.5 in. 1 ft = 4 in.

x ft.Length ratios are equal.



Course 3

7-7 Scale Drawings


0.25 x = 1 0.5 Find the cross products.

x = 2

The wall is 2 ft thick.

0.25 in. 1 ft = 0.5 in.



A. If a wall in a in. scale drawing is 0.5 in. thick, how thick is the actual wall?

14

B. How thick is the wall if a in. scale is used?


Course 3

7-7 Scale Drawings

12


0.5 x = 1 0.5 Find the cross products.

x = 1

The wall is 1 ft thick.

0.5 in. 1 ft = 0.5 in.



1. What is the scale of a drawing in which a 9 ft wall is 6 cm long?

2. Using a in. = 1 ft scale, how long would a

drawing of a 22 ft car be?

3. The height of a person on a scale drawing is

4.5 in. The scale is 1:16. What is the actual height

of the person?

The scale of a map is 1 in. = 21 mi. Find each length on the map.

4. 147 mi 5. 5.25 mi

Lesson Quiz

5.5 in.

1 cm = 1.5 ft


72 in.

7 in.

Course 3

7-7 Scale Drawings

0.25 in.

14

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