$17 \cdot 74$ In a right triangle, the lengths of the medians from the vertices of the acute angles are 5 and $\sqrt{40}$. Then the length of the hypotenuse is
(A) 10.
(B) $2 \sqrt{40}$.
(C) $\sqrt{13}$.
(D) $2 \sqrt{13}$.
(E) None of the above.
(2nd American High School Mathematics Examination, 1951)
[Solution] Let the lengths of the two legs of the right triangle be $x, y$, and the lengths of the corresponding medians be $\sqrt{40}, 5$. By the Pythagorean theorem, we have
$$
\left\{\begin{array}{l}
\left(\frac{x}{2}\right)^{2}+y^{2}=(\sqrt{40})^{2}, \\
\left(\frac{y}{2}\right)^{2}+x^{2}=5^{2} .
\end{array}\right.
$$

Adding the two equations, we get $\frac{5}{4} x^{2}+\frac{5}{4} y^{2}=65$, which simplifies to $x^{2}+y^{2}=52$. Thus, the length of the hypotenuse is $\sqrt{52}=2 \sqrt{13}$. Therefore, the answer is $(D)$.