1964 AHSME Problems/Problem 38

Problem

The sides $PQ$ and $PR$ of triangle $PQR$ are respectively of lengths $4$ inches, and $7$ inches. The median $PM$ is $3\frac{1}{2}$ inches. Then $QR$ , in inches, is:

$\textbf{(A) }6\qquad\textbf{(B) }7\qquad\textbf{(C) }8\qquad\textbf{(D) }9\qquad \textbf{(E) }10$

Solution

By the Median Formula, $PM = \frac12\sqrt{2PQ^2+2PR^2-QR^2}$

Plugging in the numbers given in the problem, we get $\frac72=\frac12\sqrt{2\cdot4^2+2\cdot7^2-QR^2}$

Solving, $7=\sqrt{2(16)+2(49)-QR^2}$ $49=32+98-QR^2$ $QR^2=81$ $QR=9=\boxed{D}$

-AOPS81619

1964 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 37	Followed by Problem 39
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1964 AHSME Problems/Problem 38

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