Question

in exercises 11 - 14, two polygons are similar. the perimeter of one polygon and the ratio of the corresponding side lengths are given. find the perimeter of the other polygon.

1. perimeter of smaller polygon: 48 cm; ratio: \\( \frac { 2 } { 3 } \\)
2. perimeter of smaller polygon: 66 ft; ratio: \\( \frac { 3 } { 4 } \\)
3. perimeter of larger polygon: 120 yd; ratio: \\( \frac { 1 } { 6 } \\)
4. perimeter of larger polygon: 85 m; ratio: \\( \frac { 2 } { 5 } \\)

Explanation:

Step1: Recall the property of similar polygons

For similar polygons, the ratio of their perimeters is equal to the ratio of their corresponding side lengths.

Step2: Solve for each problem

Problem 11

Let \(P_1 = 48\) cm (perimeter of smaller polygon) and the ratio \(r=\frac{2}{3}\). Let \(P_2\) be the perimeter of the larger polygon.
Since \(\frac{P_1}{P_2}=\frac{2}{3}\), then \(P_2=\frac{3\times P_1}{2}\).
Substitute \(P_1 = 48\) into the formula: \(P_2=\frac{3\times48}{2}=72\) cm.

Problem 12

Let \(P_1 = 66\) ft (perimeter of smaller polygon) and the ratio \(r = \frac{3}{4}\). Let \(P_2\) be the perimeter of the larger polygon.
Since \(\frac{P_1}{P_2}=\frac{3}{4}\), then \(P_2=\frac{4\times P_1}{3}\).
Substitute \(P_1 = 66\) into the formula: \(P_2=\frac{4\times66}{3}=88\) ft.

Problem 13

Let \(P_2 = 120\) yd (perimeter of larger polygon) and the ratio \(r=\frac{1}{6}\). Let \(P_1\) be the perimeter of the smaller polygon.
Since \(\frac{P_1}{P_2}=\frac{1}{6}\), then \(P_1=\frac{1\times P_2}{6}\).
Substitute \(P_2 = 120\) into the formula: \(P_1=\frac{120}{6}=20\) yd.

Problem 14

Let \(P_2 = 85\) m (perimeter of larger polygon) and the ratio \(r=\frac{2}{5}\). Let \(P_1\) be the perimeter of the smaller polygon.
Since \(\frac{P_1}{P_2}=\frac{2}{5}\), then \(P_1=\frac{2\times P_2}{5}\).
Substitute \(P_2 = 85\) into the formula: \(P_1=\frac{2\times85}{5}=34\) m.

Answer:

1. \(72\) cm
2. \(88\) ft
3. \(20\) yd
4. \(34\) m