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

(Created page with "== Problem 7== The area of a circle inscribed in an equilateral triangle is <math>48\pi</math>. The perimeter of this triangle is: <math>\textbf{(A)}\ 72\sqrt{3} \qquad \t...")
 
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<math>\textbf{(A)}\ 72\sqrt{3} \qquad \textbf{(B)}\ 48\sqrt{3}\qquad \textbf{(C)}\ 36\qquad \textbf{(D)}\ 24\qquad \textbf{(E)}\ 72  </math>   
 
<math>\textbf{(A)}\ 72\sqrt{3} \qquad \textbf{(B)}\ 48\sqrt{3}\qquad \textbf{(C)}\ 36\qquad \textbf{(D)}\ 24\qquad \textbf{(E)}\ 72  </math>   
 
==Solution==
 
==Solution==
 +
<asy>
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draw((-3,-sqrt(3))--(3,-sqrt(3))--(0,2sqrt(3))--cycle);
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draw(circle((0,0),sqrt(3)));
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dot((0,0));
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</asy>
  
 
==See Also==
 
==See Also==
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 12:06, 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)); [/asy]

See Also

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