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Understanding Ratios of Similar Polygons: Perimeters and Areas

This guide explores how to find and understand the ratios of perimeters and areas between similar polygons, using examples to illustrate the concepts. We begin with the ratio of the perimeters, then move on to the areas, referring to relevant theorems for clarity. Through guided practice and standardized test examples, we demonstrate how to derive these ratios and solve problems effectively. By applying these principles, you can easily calculate cost estimates based on area ratios, making this knowledge practical and applicable.

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Understanding Ratios of Similar Polygons: Perimeters and Areas

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  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. ANSWER 16 ; 36 ft2 9 for Examples 1 and 2 GUIDED PRACTICE 1. The perimeter of ∆ABC is 16 feet, and its area is 64 square 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.

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