EXAMPLE 1

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

2. 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 22:32, or 4:9. The ratio of the lengths of corresponding sides is 2 8 , or2:3. = 12 3 EXAMPLE 1 Find ratios of similar polygons SOLUTION

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 72:52, or 49:25. This ratio is also the ratio of the carpeting costs. Let xbe the cost for the den.

4. x 49 cost of carpet for den = 25 225 cost of carpet for bedroom = x441 ANSWER It costs $ 441 to carpet the den. The correct answer is D. EXAMPLE 2 Standardized Test Practice Solve for x.

5. F B SOLUTION D A C E The ratio of the lengths of the perimeters is 4 16 , or4:3. = 12 3 for Examples 1 and 2 GUIDED PRACTICE 1. The perimeter of ∆ABC is 16 feet, and its area is 64 feet. The perimeter of ∆DEF is 12feet. Given ∆ABC ≈∆DEF, find the ratio of the area of ∆ABC to the area of ∆DEF. Then find the area of ∆DEF.

6. By Theorem 11.7, the ratio of the areas is 42:32, or 16:9, 16 9 Let x be the area of ∆ DEF 16 64 = 9 x 16 x = 64 9 ∆ Area of = 36 ft2 DEF for Examples 1 and 2 GUIDED PRACTICE Write Proportion Cross Product Property x = 36 Solve for x