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:
$\textbf{(A) }6\qquad\textbf{(B) }7\qquad\textbf{(C) }8\qquad\textbf{(D) }9\qquad \textbf{(E) }10$
By the [Median Formula](https://artofproblemsolving.com/wiki/index.php?title=Median_Formula&action=edit&redlink=1), $PM = \frac12\sqrt{2PQ^2+2PR^2-QR^2}$
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