Difference between revisions of "1953 AHSME Problems/Problem 37"

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{{AHSME 50p box|year=1953|num-b=36|num-a=38}}
 
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Revision as of 20:40, 24 January 2020

Problem

The base of an isosceles triangle is $6$ inches and one of the equal sides is $12$ inches. The radius of the circle through the vertices of the triangle is:

$\textbf{(A)}\ \frac{7\sqrt{15}}{5} \qquad \textbf{(B)}\ 4\sqrt{3} \qquad \textbf{(C)}\ 3\sqrt{5} \qquad \textbf{(D)}\ 6\sqrt{3}\qquad \textbf{(E)}\ \text{none of these}$

Solution

See Also

1953 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 36
Followed by
Problem 38
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