1953 AHSME Problems/Problem 37

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

1953 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 36	Followed by Problem 38
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1953 AHSME Problems/Problem 37

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