3-5 HW: Pg. 165-167 #6-42eoe, 50-51, 56-57. 56. C 57. D

3-5 HW: Pg. 165-167 #6-42eoe, 50-51, 56-57

56. C 57. D

Wir2.5• LESSON

3-6 Solve Proportions Using Cross Products

Cross Product: __________________________________

________________________

Scale Drawing: __________________________________________

_______________________________________

denominator of the other ratio.

2dimensional drawing of an object where dimensions

product of numerator of 1 ratio and the

of drawing are in proportion to the real object

Use the cross products property

EXAMPLE 1

8 15 = x 6

Solve the proportion = .8 x

615

Cross products property

Simplify.120 = 6x

Divide each side by 6.20 = x

The solution is 20. Check by substituting 20 for x in the original proportion.

ANSWER

Substitute 20 for x.CHECK

Cross products property

Simplify. Solution checks.

820

615

=?

8 15 = 20 6?

120 = 120

EXAMPLE 2 Standardized Test Practice

What is the value of x in the proportion = ?4x

83x –

A – 6 B – 3 C 3 D 6

SOLUTION

4x =

8x – 3

Write original proportion.

Cross products property4(x – 3) = x 8

4x – 12 = 8x Simplify.

Subtract 4x from each side. –12 = 4x

Divide each side by 4. –3 = x

The value of x is –3. The correct answer is B.

ANSWER

A B C D

Write and solve a proportion

EXAMPLE 3

x280

= 8 amount of food100 weight of seal

SOLUTION

STEP 1 Write a proportion involving two ratios that compare the amount of food with the weight of the seal.

Each day, the seals at an aquarium are each fed 8 pounds of food for every 100 pounds of their body weight. A seal at the aquarium weighs 280 pounds.How much food should the seal be fed per day?

Seals

STEP 2 Solve the proportion.

8 280 = 100 x Cross products property

2240 = 100x Simplify.

22.4 = x Divide each side by 100.

ANSWER A 280 pound seal should be fed 22.4 pounds of food per day.

EXAMPLE 1

Solve the proportion. Check your solution.

GUIDED PRACTICE for Examples 1, 2, and 3

=4 a

2430

1. 5ANSWER

3x =

2x – 6

2.18ANSWER

4m5 =

m – 6 3. 30ANSWER

20.8ANSWER

WHAT IF? In Example 3, suppose the seal weighs 260 pounds. How much food should the seal be fed per day?

4.

EXAMPLE 4 Use the scale on a map

Maps Use a metric ruler and the map of Ohio to estimate the distance between Cleveland and Cincinnati.

SOLUTION

From the map’s scale, 1 centimeter represents 85 kilometers. On the map, the distance between Cleveland and Cincinnati is about 4.2 centimeters.

Write and solve a proportion to find the distance d between the cities.

=3

d 1 centimeters85 kilometers

Cross products property

d = 255 Simplify.ANSWER The actual distance between Cleveland and

Cincinnati is about 255 kilometers.

1 d = 85 3

EXAMPLE 4 Use the scale on a mapGUIDED PRACTICE for Example 4

5.

Use a metric ruler and the map in Example 4 to estimate the distance (in kilometers) between Columbus and Cleveland. about 144.5 kmANSWER

6.

The ship model kits sold at a hobby store have a scale of 1 ft : 600 ft. A completed model of the Queen Elizabeth II is 1.6 feet long. Estimate the actual length of the Queen Elizabeth II.

Model ships

ANSWER about 960 ft

Summary• How do you solve proportions using cross products?• Ans: Cross products of a proportion are equal. Multiply

the numerator of each ratio by the denominator of the other ratio and write an equal sign in between the 2 products. Then solve for the variable.

• If the distance between Columbus and Cincinnati is about 170 km, and the map’s scale is drawn so 1 cm = 85 km. Can you estimate the distance between the two cities?

• Ans: 1

85 17085 170

= 2, so distance between Columbus

and Cincinnati is about 2 kil

8

omete s.

5

r

85

x

x

x

Check Yourself

Pg. 171-173 #4-24eoe, 33-34, 44-45 and

Quiz Pg. 173 #2-16e