# 1957 AHSME Problems/Problem 7

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## Problem 7

The area of a circle inscribed in an equilateral triangle is $48\pi$. The perimeter of this triangle is: $\textbf{(A)}\ 72\sqrt{3} \qquad \textbf{(B)}\ 48\sqrt{3}\qquad \textbf{(C)}\ 36\qquad \textbf{(D)}\ 24\qquad \textbf{(E)}\ 72$

## Solution $[asy] draw((-3,-sqrt(3))--(3,-sqrt(3))--(0,2sqrt(3))--cycle); draw(circle((0,0),sqrt(3))); dot((0,0)); draw((0,0)--(0,-sqrt(3))); [/asy]$ We can see that the radius of the circle is $4\sqrt{3}$. We know that the radius is $\frac{1}{3}$ of each median line of the triangle; each median line is therefore $12\sqrt{3}$. Since the median line completes a $30$- $60$- $90$ triangle, we can conclude that one of the sides of the triangle is $24$. Triple the side length and we get our answer, $\boxed{\textbf{(E)}72}$.

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