An isosceles ΔMNP is inscribed in a circle. If MN = MP = 16√5 cm, and NP = 32 cm, what is the radius (in cm) of the circle?

Answer: Option A

Solution

Answer: Option A

Solution:

$$\begin{array}{rl}& \text{Height of }\Delta \text{MND}\\ & =\sqrt{{\left(16\sqrt{5}\right)}^2-{\left(\frac{32}{2}\right)}^2}\\ & =\sqrt{{\left(16\sqrt{5}\right)}^2-{\left(16\right)}^2}\\ & =16\sqrt{5-1}\\ & =16\times 2\\ & =32\\ & \text{R}=\frac{\text{abc}}{4\Delta }=\frac{16\sqrt{5}\times 16\sqrt{5}\times 32}{4\times \frac{1}{2}\times 32\times 32}=20\end{array}$$