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\sqrt{21} inches
16\sqrt{21} inches
21 inches

Explanation:

Step1: Divide the base of the isosceles - triangle

Since the base of each isosceles triangle is 16 inches, when we consider the right - triangles formed by the height and half of the base of the isosceles triangle, the length of half of the base $b = \frac{16}{2}=8$ inches.

Step2: Calculate the heights of the two isosceles triangles using the Pythagorean theorem

For the first isosceles triangle with leg length $a_1 = 10$ inches, let the height be $h_1$. By the Pythagorean theorem $h_1=\sqrt{a_1^{2}-b^{2}}=\sqrt{10^{2}-8^{2}}=\sqrt{100 - 64}=\sqrt{36}=6$ inches.
For the second isosceles triangle with leg length $a_2 = 17$ inches, let the height be $h_2$. By the Pythagorean theorem $h_2=\sqrt{a_2^{2}-b^{2}}=\sqrt{17^{2}-8^{2}}=\sqrt{289 - 64}=\sqrt{225}=15$ inches.

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

The length of the other diagonal of the kite is the sum of the heights of the two isosceles triangles, $h = h_1+h_2=6 + 15=21$ inches.

Answer:

D. 21 inches