# Use the ratios of perimeter and area to solve problems involving polygons.

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RBcryy

Mathematics

## Description

Directions: Solve each of the following problems.

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1. Two A are similar. The sides of the first A are 2, 4, and 6. The largest side of the second A is 24. Find the perimeter of the second A. 2. The areas of two similar polygons are in the ratio 64:81. Find the ratio of the corresponding sides. 3. Finding the areas of similar right triangles whose scale factor is 3: 5. 8 10 10 12 M 8 N Х 6 Y 4. The perimeters of two similar triangles is in the ratio 2: 4. The sum of their areas is 100 cm². Find the area of each triangle. 5. Two A are similar. The sides of the first A are 4, 5, and 6. The largest side of the second A is 24. Find the perimeter of the second A. 6. The areas of two similar polygons are in the ratio 36:16. Find the ratio of the corresponding sides. 7. Finding the areas of similar right triangles whose scale factor is 3: 1. 12 6 9 Р 3 A 6 B 8. Two A are similar. The sides of the first A are 5, 10, and 15. The largest side of the second A is 20. Find the perimeter of the second A. 9. The areas of two similar polygons are in the ratio 36:49. Find the ratio of the corresponding sides. 10. The areas of two similar polygons are in the ratio 121:100. Find the ratio of the corresponding sides.
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The solutions are ready. Please note that the questions 3 has only 1 right triangle of two, while the question 7 has no right triangles. I had to find their areas using Heron's formula.

24

1. The largest side of the first triangle is 6, thus the second triangle is

6

= 4 times greater.

This way the perimeter of the second triangle is (2 + 4 + 6) ∙ 4 = 𝟒𝟖.
[note that 2, 4, 6 is not a so good triangle]

2. The ratio of areas is the ratio of sides squared.
Thus, the ratio of sides is √64: √81 = 𝟖: 𝟗.

3. The area of the first triangle is

1
2

∙ 8 ∙ 6 = 24.
5 2

𝟐𝟎𝟎

If a triangle has a ratio 3: 5 to the given, its area is 24 ∙ ( ) =
3

𝟑

.

The second triangle is not right because 102 + 82 ≠ 122 .
1
Use Heron’s formula: 𝑠 = 2 (8 + 10 + 12) = 15, the area is √15 ∙ 7 ∙ 5 ∙ 3 = 15√7.
5 2

𝟏𝟐𝟓√𝟕

If a triangle has a ...

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