1958 AHSME Problems/Problem 16

Problem

The area of a circle inscribed in a regular hexagon is $100\pi$ . The area of hexagon is:

$\textbf{(A)}\ 600\qquad \textbf{(B)}\ 300\qquad \textbf{(C)}\ 200\sqrt{2}\qquad \textbf{(D)}\ 200\sqrt{3}\qquad \textbf{(E)}\ 120\sqrt{5}$

Solution

We can split the hexagon into 6 equilateral triangles. If the area of the circle is $100pi$ , then the radius is $10$ . The radius is equal to the height of one of the equilateral triangles. Using $sin60 = \frac{\sqrt3}{2}$ , we get the hypotenuse is $\frac{20\sqrt{3}}{3}$ , which is also equal to the side length. Using the hexagon formula (or going back to the equilateral triangle formula $\cdot 6$ ), we get

$\frac{6}{4} \cdot s^2\sqrt{3} \Rightarrow$ $\frac{6}{4} \cdot (\frac{20\sqrt3}{3})^2 \cdot \sqrt3 \Rightarrow$ $\frac{6}{4} \cdot \frac{20\sqrt3}{3} \cdot \frac{20\sqrt3}{3} \cdot \sqrt3 \Rightarrow$ $\boxed{200\sqrt3}$

1958 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 15	Followed by Problem 17
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1958 AHSME Problems/Problem 16

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