1. Properties of Proportions 7-2

2. EXAMPLE 4 Use a scale drawing SOLUTION Maps The scale of the map at the right is 1 inch : 26 miles. Find the actual distance from Pocahontas to Algona. Use a ruler. The distance from Pocahontas to Algona on the map is about 1.25 inches. Let x be the actual distance in miles.

3. EXAMPLE 4 Use a scale drawing Cross Products Property Simplify. = 1.25 (26) x x = 32.5 The actual distance from Pocahontas to Algona is about 32.5 miles. 1.25 in. x mi 26 mi 1 in. = distance on map actual distance

4. EXAMPLE 5 Solve a multi-step problem What is the diameter of the actual dome? a. b. About how many times as tall as your model is the actual building? Scale Model You buy a 3-D scale model of the Reunion Tower in Dallas, TX. The actual building is 560 feet tall. Your model is 10 inches tall, and the diameter of the dome on your scale model is about 2.1 inches.

5. EXAMPLE 5 Solve a multi-step problem Cross Products Property measurement on model measurement on actual building = 1176 10 x x = 117.6 SOLUTION 10 in. 560 ft x ft 2.1 in. = a. Solve for x. The diameter of the actual dome is about 118 feet. ANSWER

6. EXAMPLE 5 Solve a multi-step problem b. To simplify a ratio with unlike units, multiply by a conversion factor. 560 ft 10 in. = 672 The actual building is 672 times as tall as the model. ANSWER 1 ft 12 in. 560 ft 10 in. =

7. GUIDED PRACTICE for Examples 4 and 5 4. Two cities are 96 miles from each other. The cities are 4 inches apart on a map. Find the scale of the map. SOLUTION The distance between two cities are 96 miles, and the distance on the map is 4 inches distance on map actual distance 96 mi 4 in = = 1 24 Simplify. = 1 : 24

8. GUIDED PRACTICE for Examples 4 and 5 ANSWER Scale of the map is 1 : 24 = 1 in : 24 mi

9. GUIDED PRACTICE for Examples 4 and 5 5. What If ? Your friend has a model of the Reunion Tower that is 14 inches tall. What is the diameter of the dome on your friend’s model? = 1646.4 560 x x = 2.94 SOLUTION 14 in. 560 ft 117.6 ft x in. = a. ANSWER 2.94 in. Cross Products Property measurement on model measurement on actual building Solve for x.

10. EXAMPLE 1 Use properties of proportions SOLUTION NP ST MN RS = Because 4 x = 8 10, then In the diagram, NP ST MN RS = Write four true proportions. By the Reciprocal Property, the reciprocals are equal, so x 4 = 10 8 By Property 3, you can interchange the means, so = 8 4 10 x

11. EXAMPLE 1 Use properties of proportions By Property 4, you can add the denominators to the numerators, so 8 + 10 10 4 + x x x =, or 18 10 =.

12. EXAMPLE 2 Use proportions with geometric figures Given Property of Proportions (Property 4 ) Substitution Property of Equality Cross Products Property Solve for x. BE EC BD DA = SOLUTION 18 + 6 6 = x 3 x 12 = 6x6x3(18 + 6) = So, BA = 12 and BD = 12 – 3 = 9. ALGEBRA In the diagram, BE EC BD DA = Find BA and BD. = BE + EC EC BD + DA DA

13. EXAMPLE 3 Find the scale of a drawing SOLUTION To find the scale, write the ratio of a length in the drawing to an actual length, then rewrite the ratio so that the denominator is 1. The scale of the blueprint is 2.5 cm : 1 cm. = 5 cm 2 cm = 5 2 2 = 2.5 1 The blueprint shows a scale drawing of a cell phone. The length of the antenna on the blueprint is 5 centimeters. The actual length of the antenna is 2 centimeters. What is the scale of the blueprint? Blueprints length on blueprint length of antenna

14. GUIDED PRACTICE for Examples 1, 2 and 3 1. In Example 1, find the value of x. RS ST MN NP = SOLUTION Write proportion 10 x = 8 4 Substitute. = 8 x4 10 8x8x =40 Cross Product property Multiply x =5 Divide ANSWER The value of x is 5

15. GUIDED PRACTICE for Examples 1, 2 and 3 2. In Example 2, find the value of x. SOLUTION BE BC DE AC = Given Property of Proportions (Property 4 ) Substitution Property of Equality Cross Products Property Solve for AC. BE BC DE AC = = BE + BC BC DE + AC AC 16= = 12 + AC AC 18 + (18 + 6) 18 + 6 = 22(12 + AC)42 AC

16. GUIDED PRACTICE for Examples 1, 2 and 3 ANSWER So, AC = 16

17. GUIDED PRACTICE for Examples 1, 2 and 3 3. What If ? In Example 3, suppose the length of the antenna on the blueprint is 10 centimeters. Find the new scale of the blueprint. SOLUTION To find the scale, write the ratio of a length in the drawing to an actual length, then rewrite the ratio so that the denominator is 1. The scale of the blueprint is 5 cm : 1 cm. length on blueprint length of antenna = 10 cm 2 cm 10 2 = 2 = 5 1