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