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 kite’s other diagonal?
o 6 inches
o $4\sqrt{21}$ inches
o $16\sqrt{21}$ inches
o 21 inches

Explanation:

Step1: Divide the base of the isosceles triangle

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

Step2: Calculate the height of the first isosceles triangle

Let the legs of the isosceles triangles be $a = 10$ inches and $b = 17$ inches. For the first isosceles triangle with leg length $a = 10$ inches, using the Pythagorean theorem $h_1=\sqrt{a^{2}-(\frac{16}{2})^{2}}=\sqrt{10^{2}-8^{2}}=\sqrt{100 - 64}=\sqrt{36}=6$ inches.

Step3: Calculate the height of the second isosceles triangle

For the second isosceles triangle with leg length $b = 17$ inches, using the Pythagorean theorem $h_2=\sqrt{b^{2}-(\frac{16}{2})^{2}}=\sqrt{17^{2}-8^{2}}=\sqrt{289 - 64}=\sqrt{225}=15$ inches.

Step4: Find 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, so $h=h_1 + h_2=6 + 15=21$ inches.

Answer:

21 inches