Difference between revisions of "1957 AHSME Problems/Problem 7"

Revision as of 11:20, 12 October 2020

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, $\boxed{\textbf{(E)}72}$ .

1957 AHSME (Problems • Answer Key • Resources)
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All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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+{{AHSME box|year=1957|num-b=6|num-a=8}}
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Difference between revisions of "1957 AHSME Problems/Problem 7"

Revision as of 11:20, 12 October 2020

Problem 7

Solution

See Also