1. EXAMPLE 1 Find ratios of similar polygons Ratio (red to blue) of the perimeters a.a. Ratio (red to blue) of the areas b.b. In the diagram, ABC  DEF. Find the indicated ratio. ∆∆

2. EXAMPLE 1 Find ratios of similar polygons a. By Theorem 6.1 on page 374, the ratio of the perimeters is 2:3. b. By Theorem 11.7 above, the ratio of the areas is 2 2 :3 2, or 4:9. SOLUTION The ratio of the lengths of corresponding sides is 12 8 = 3 2, or 2:3.

3. EXAMPLE 2 Standardized Test Practice SOLUTION The ratio of a side length of the den to the corresponding side length of the bedroom is 14 : 10, or 7 : 5. So, the ratio of the areas is 7 2 : 5 2, or 49 : 25. This ratio is also the ratio of the carpeting costs. Let x be the cost for the den.

4. EXAMPLE 2 Standardized Test Practice 25 49 = 225 x cost of carpet for den cost of carpet for bedroom x 441 = Solve for x. ANSWER It costs $ 441 to carpet the den. The correct answer is D.

5. 1. The perimeter of ∆ABC is 16 feet, and its area is 64 feet. The perimeter of ∆DEF is 12 feet. Given ∆ABC ≈ ∆DEF, find the ratio of the area of ∆ABC to the area of ∆DEF. Then find the area of ∆DEF. GUIDED PRACTICE for Examples 1 and 2 SOLUTION A B C E D F The ratio of the lengths of the perimeters is 12 16 = 3 4, or 4:3.

6. GUIDED PRACTICE for Examples 1 and 2 By Theorem 11.7, the ratio of the areas is 4 2 :3 2, or 16:9, 9 16 DEF Let x be the area of ∆ 9 16 x 64 = 16 x= 9 64 x = 36 Write Proportion Cross Product Property Solve for x Area of = 36 ft 2 DEF ∆

7. EXAMPLE 3 Use a ratio of areas Cooking SOLUTION First draw a diagram to represent the problem. Label dimensions and areas. Then use Theorem 11.7. If the area ratio is a 2 : b 2, then the length ratio is a:b. A large rectangular baking pan is 15 inches long and 10 inches wide. A smaller pan is similar to the large pan. The area of the smaller pan is 96 square inches. Find the width of the smaller pan.

8. EXAMPLE 3 Use a ratio of areas Area of smaller pan Area of large pan = 150 96 = 25 16 Write ratio of known areas. Then simplify. = 4 5 Find square root of area ratio. Length in smaller pan Length in large pan ANSWER Any length in the smaller pan is, or 0.8, of the corresponding length in the large pan. So, the width of the smaller pan is 0.8(10 inches ) 8 inches. 4 5 =

9. EXAMPLE 4 Solve a multi-step problem The floor of the gazebo shown is a regular octagon. Each side of the floor is 8 feet, and the area is about 309 square feet. You build a small model gazebo in the shape of a regular octagon. The perimeter of the floor of the model gazebo is 24 inches. Find the area of the floor of the model gazebo to the nearest tenth of a square inch. Gazebo

10. EXAMPLE 4 SOLUTION All regular octagons are similar, so the floor of the model is similar to the floor of the full-sized gazebo. STEP 1 Find the ratio of the lengths of the two floors by finding the ratio of the perimeters. Use the same units for both lengths in the ratio. Perimeter of full-sized Perimeter of model = 8(8 ft ) 24 in. = 64 ft 2 ft = 32 1 So, the ratio of corresponding lengths (full-sized to model) is 32:1. Solve a multi-step problem

11. EXAMPLE 4 Solve a multi-step problem STEP 2 (length of full-sized) 2 (length of model) 2 Calculate the area of the model gazebo’s floor. Let x be this area. Area of full-sized Area of model = 309 ft 2 x ft 2 32 2 1212 = Theorem 11.7 Substitute. 1024x 309= Cross Products Property x ≈ 0.302 ft 2 Solve for x.

12. EXAMPLE 4 Solve a multi-step problem STEP 3 Convert the area to square inches. 0.302 ft 2 144 in. 2 1 ft. 2 ≈ 43. 5 in. 2 The area of the floor of the model gazebo is about 43.5 square inches. ANSWER

13. GUIDED PRACTICE for Examples 3 and 4 2. The ratio of the areas of two regular decagons is 20:36. What is the ratio of their corresponding side lengths in simplest radical form? SOLUTION Then use Theorem 11.7. If the area ratio is a 2 : b 2, then the length ratio is a:b. Area of smaller decagon Area of large decagon = 9 5 Write ratio of known areas. Then simplify. Find square root of area as ratios. Length in smaller decagon Length in large decagon = 36 20 = 5 3 √

14. GUIDED PRACTICE for Examples 3 and 4 3. Rectangles I and II are similar. The perimeter of Rectangle I is 66 inches. Rectangle II is 35 feet long and 20 feet wide. Show the steps you would use to find the ratio of the areas and then find the area of Rectangle I. Perimeter of Rectangle I = 66 inches SOLUTION Perimeter of Rectangle II = 2(l + b) = 2(35 + 20) = 2 35 + 2 20 = 110 feet.

15. GUIDED PRACTICE for Examples 3 and 4 STEP 1 Find the ratio of the lengths of the two rectangles by finding the ratio of the perimeters. = 66 1320 = 1 20 Perimeter of Rectangle I Perimeter of Rectangle II So, the ratio of the corresponding lengths to is 1:20. Convert 110 feet to inches. 110 ft= 110 12 = 1320 inches.

16. GUIDED PRACTICE for Examples 3 and 4 STEP 2 STEP 3 Find the area of Rectangle II. Area of Rectangle II = lb = 35(20) = 700 feet 2 Convert 700 feet 2 to inches 2. 700 ft 2 = 700 144 = 100,800 inches 2. Find the ratio of their Areas (sides of Rectangle I ) 2 (sides of Rectangle II ) 2 = Area of Rectangle I Area of Rectangle II 1 20 2 = 1 400

17. GUIDED PRACTICE for Examples 3 and 4 STEP 4 Find the area of Rectangle I. Area of Rectangle I = Area of Rectangle II · (ratio of areas) = 252 in. 2 Area of Rectangle I = 100,800 · 1 400