Question

charlie puts together two isosceles triangles so that they share a base, creating a kite. the legs of the triangles are 10 inches and 17 inches, respectively. if the length of the base for both triangles is 16 inches long, what is the length of the kites other diagonal?
6 inches
4\sqrt{11} inches
14\sqrt{11} inches
21 inches

Explanation:

Step1: Find the height of the first isosceles triangle

Use the Pythagorean theorem \(a^{2}+b^{2}=c^{2}\). For the first isosceles triangle with leg \(c = 10\) inches and half - base \(a=\frac{16}{2}=8\) inches. Let the height be \(h_1\). Then \(h_1=\sqrt{10^{2}-8^{2}}=\sqrt{100 - 64}=\sqrt{36}=6\) inches.

Step2: Find the height of the second isosceles triangle

For the second isosceles triangle with leg \(c = 17\) inches and half - base \(a = 8\) inches. Let the height be \(h_2\). Then \(h_2=\sqrt{17^{2}-8^{2}}=\sqrt{289-64}=\sqrt{225}=15\) inches.

Step3: Calculate the length of the other diagonal of the kite

The length of the other diagonal of the kite \(d=h_1 + h_2\). So \(d=6 + 15=21\) inches.

Answer:

21 inches