# 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 $[asy] draw((0,0)--(3,3sqrt(15))--(6,0)--cycle); draw((3,3sqrt(15))--(3,0)); label("A",(3,3sqrt(15)),N); label("B",(0,0),SW); label("C",(6,0),SE); label("D",(3,0),S); label("12",(1.5,5.8),WNW); label("12",(4.5,5.8),ENE); label("3",(1.5,0),N); [/asy]$ Let $\triangle ABC$ be an isosceles triangle with $AB=AC=12,$ and $BC = 6$. Draw altitude $\overline{AD}$. Since $A$ is the apex of the triangle, the altitude $\overline{AD}$ is also a median of the triangle. Therefore, $BD=CD=3$. Using the Pythagorean Theorem, $AD=\sqrt{12^2-3^2}=3\sqrt{15}$. The area of $\triangle ABC$ is $\frac12\cdot 6\cdot 3\sqrt{15}=9\sqrt{15}$.

If $a,b,c$ are the sides of a triangle, and $A$ is its area, the circumradius of that triangle is $\frac{abc}{4A}$. Using this formula, we find the circumradius of $\triangle ABC$ to be $\frac{6\cdot 12\cdot 12}{4\cdot 9\sqrt{15}}=\frac{2\cdot 12}{\sqrt{15}}=\frac{24\sqrt{15}}{15}=\frac{8\sqrt{15}}{5}$. The answer is $\boxed{\textbf{(E)}\ \text{none of these}}$

## See Also

 1953 AHSC (Problems • Answer Key • Resources) Preceded byProblem 36 Followed byProblem 38 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50 All AHSME Problems and Solutions

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