$17 \cdot 96$ In $\triangle A B C$, the median from $A$ is perpendicular to the median from vertex $B$. If the sides $A C$ and $B C$ are 6 and 7 respectively, then the length of $A B$ is
(A) $\sqrt{17}$.
(B) 4 .
(C) $4 \frac{1}{2}$.
(D) $2 \sqrt{5}$.
(E) $4 \frac{1}{4}$.
(12th American High School Mathematics Examination, 1961;
21st American High School Mathematics Examination, 1970)
[Solution] Let the intersection point of the medians $A M, B N$ be the centroid $G$.
Obviously, if we set $G M=u, G N=v$, then $A G=2 u, B G=2 v$. According to the given, $B M=\frac{1}{2} B C=3 \frac{1}{2}$,
$$
A N=\frac{1}{2} A C=3 \text {. }
$$

By the Pythagorean theorem, in the right triangles $\triangle A G N$ and $\triangle B G M$, we have $(2 u)^{2}+v^{2}=3^{2}$ and $(2 v)^{2}+u^{2}=\left(3 \frac{1}{2}\right)^{2}$.
Adding these two equations, we get
$$
5 u^{2}+5 v^{2}=\frac{85}{4} \text {. }
$$

Thus, $4 u^{2}+4 v^{2}=17=A B^{2}$. Therefore, $A B=\sqrt{17}$. Hence, the answer is $(A)$.