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

See Also

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