$17 \cdot 85 \triangle P Q R$ has sides $P Q, P R$ of lengths 4 and 7, respectively, and the median $P M$ has a length of $3 \frac{1}{2}$. Then the length of $Q R$ is 
(A) 6.
(B) 7.
(C) 8.
(D) 9.
(E) 10.
(15th American High School Mathematics Examination, 1964)
[Solution] Let $M R=y$. Draw $P H \perp Q R$ at $H$, and let $H M=x$. By the Pythagorean theorem,
$$
16-(y-x)^{2}=\left(3 \frac{1}{2}\right)^{2}-x^{2}
$$

which simplifies to $(y-x)^{2}-x^{2}=\frac{15}{4}$, or $y^{2}-2 x y=\frac{15}{4}$.
Also, $16-(y-x)^{2}=49-(x+y)^{2}$,

which simplifies to $(y+x)^{2}-(y-x)^{2}=33$, or $4 x y=33$. From the above two equations, we have $y^{2}=\frac{33}{2}+\frac{15}{4}=\frac{81}{4}$, giving $y=\frac{9}{2}$. Thus, $Q R=9$. Therefore, the answer is $(D)$.