17.98 $A B$ is the hypotenuse of the right triangle $A B C$, the median $A D$ is 7, and the median $B E$ is 4, then the length of $A B$ is
(A) 10.
(B) $5 \sqrt{3}$.
(C) $5 \sqrt{2}$.
(D) $2 \sqrt{13}$.
(E) $2 \sqrt{15}$.
(9th American High School Mathematics Examination, 1958)
［Solution］From the given and Pythagorean theorem, we have
$$
\begin{array}{l}
7^{2}=C D^{2}+A C^{2}, \\
4^{2}=B C^{2}+C E^{2},
\end{array}
$$

Thus, $7^{2}+4^{2}=\left(C D^{2}+B C^{2}\right)+\left(A C^{2}+\right.$ $\left(E^{2}\right)$.

From the given, then
$$
\begin{array}{c}
C D=\frac{1}{2} B C, \quad C E=\frac{1}{2} A C, \\
7^{2}+4^{2}=\frac{5}{4}\left(B C^{2}+A C^{2}\right)=\frac{5}{4} A B^{2} .
\end{array}
$$

Therefore, $A B^{2}=\frac{4}{5}\left(7^{2}+4^{2}\right)=52$, which gives $A B=2 \sqrt{13}$. Hence, the answer is $(D)$.