1962 AHSME Problems/Problem 39


Two medians of a triangle with unequal sides are $3$ inches and $6$ inches. Its area is $3 \sqrt{15}$ square inches. The length of the third median in inches, is: $\textbf{(A)}\ 4\qquad\textbf{(B)}\ 3\sqrt{3}\qquad\textbf{(C)}\ 3\sqrt{6}\qquad\textbf{(D)}\ 6\sqrt{3}\qquad\textbf{(E)}\ 6\sqrt{6}$


By the area formula: \[A = \frac43\sqrt{s(s-m_1)(s-m_2)(s-m_3)}\] Where $s = \frac{m_1+m_2+m_3}{2}$. Plugging in the numbers: \[3\sqrt{15} = \frac43\sqrt{\frac{9+m}2\cdot\frac{9-m}2\cdot\frac{m+3}2\cdot\frac{m-3}2}\] Simplifying and squaring both sides: \[1215 = (9-m)(9+m)(m+3)(m-3)\] Now, we can just plug in the answer choices and find that $\boxed{3\sqrt6}$ works.

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