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

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draw(circle((0,0),sqrt(3)));
 
draw(circle((0,0),sqrt(3)));
 
dot((0,0));
 
dot((0,0));
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draw((0,0)--(0,-sqrt(3)));
 
</asy>
 
</asy>
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We can see that the radius of the circle is <math>4\sqrt{3}</math>. We know that the radius is <math>\frac{1}{3}</math> of each median line of the triangle; each median line is therefore <math>12\sqrt{3}</math>. Since the median line completes a <math>30</math>-<math>60</math>-<math>90</math> triangle, we can conclude that one of the sides of the triangle is <math>24</math>. Triple the side length and we get our answer, <math>\boxed{\textbf{(E)}72}</math>.
  
 
==See Also==
 
==See Also==
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{{AHSME 50p box|year=1957|num-b=6|num-a=8}}
 
{{MAA Notice}}
 
{{MAA Notice}}
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[[Category:Introductory Geometry Problems]]

Latest revision as of 08:07, 25 July 2024

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

See Also

1957 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 6 		Followed by
Problem 8
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All AHSME Problems and Solutions

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