Question

charlene 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√21 inches 16√21 inches 21 inches

Explanation:

Step1: Calculate the height of the first isosceles triangle

The base of the isosceles triangle is \(16\) inches. The half - base is \(\frac{16}{2}=8\) inches. The leg of the first isosceles triangle is \(10\) inches. Using the Pythagorean theorem \(a^{2}+b^{2}=c^{2}\), where \(c = 10\) (hypotenuse) and \(a = 8\) (one of the legs of the right - triangle formed by half - base and height). Let the height of the first triangle be \(h_1\). Then \(h_1=\sqrt{10^{2}-8^{2}}=\sqrt{100 - 64}=\sqrt{36}=6\) inches.

Step2: Calculate the height of the second isosceles triangle

The half - base is still \(8\) inches. The leg of the second isosceles triangle is \(17\) inches. Let the height of the second triangle be \(h_2\). Using the Pythagorean theorem \(h_2=\sqrt{17^{2}-8^{2}}=\sqrt{289 - 64}=\sqrt{225}=15\) inches.

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

The length of the other diagonal of the kite (formed by the sum of the heights of the two isosceles triangles) is \(h_1 + h_2\). So, \(6+15 = 21\) inches.

Answer:

21 inches