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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. ∆∆

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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.

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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.

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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.

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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.

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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 ∆

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EXAMPLE 3 Standardized Test Practice.

EXAMPLE 3 Standardized Test Practice.

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