Question

in exercises 17 - 20, the polygons are similar. the area of one polygon is given. find the area of the other polygon. example 5 17. 3 ft a = 27 ft² 6 ft 18. 4 cm a = 10 cm² 12 cm 19. 4 in. 20 in. a = 100 in.² 20. 3 cm 12 cm a = 96 cm²

Explanation:

Step1: Find the ratio of the corresponding sides

For exercise 17, the ratio of the sides is \(\frac{6}{3}=2\).
For exercise 18, the ratio of the sides is \(\frac{12}{4}=3\).
For exercise 19, the ratio of the sides is \(\frac{20}{4}=5\).
For exercise 20, the ratio of the sides is \(\frac{12}{3}=4\).

Step2: Use the ratio of areas formula

The ratio of the areas of two similar polygons is the square of the ratio of their corresponding sides. Let the area of the unknown polygon be \(A_{2}\), the area of the known polygon be \(A_{1}\), and the ratio of the sides be \(k\). Then \(A_{2}=A_{1}\times k^{2}\) (if \(A_{2}\) is larger) or \(A_{2}=\frac{A_{1}}{k^{2}}\) (if \(A_{2}\) is smaller).

• Exercise 17:

Since \(k = 2\) and \(A_{1}=27\), \(A_{2}=27\times2^{2}=27\times4 = 108\) \(ft^{2}\).

• Exercise 18:

Since \(k = 3\) and \(A_{1}=10\), \(A_{2}=10\times3^{2}=10\times9 = 90\) \(cm^{2}\).

• Exercise 19:

Since \(k = 5\) and \(A_{1}=100\), \(A_{2}=\frac{100}{5^{2}}=\frac{100}{25}=4\) \(in^{2}\).

• Exercise 20:

Since \(k = 4\) and \(A_{1}=96\), \(A_{2}=\frac{96}{4^{2}}=\frac{96}{16}=6\) \(cm^{2}\).

Answer:

1. \(108\) \(ft^{2}\)
2. \(90\) \(cm^{2}\)
3. \(4\) \(in^{2}\)
4. \(6\) \(cm^{2}\)