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

See Also

1964 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 37
Followed by
Problem 39
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