$\overline{AB}$ is the hypotenuse of a right triangle $ABC$. Median $\overline{AD}$ has length $7$ and median $\overline{BE}$ has length $4$. The length of $\overline{AB}$ is:
$\textbf{(A)}\ 10\qquad  \textbf{(B)}\ 5\sqrt{3}\qquad  \textbf{(C)}\ 5\sqrt{2}\qquad  \textbf{(D)}\ 2\sqrt{13}\qquad  \textbf{(E)}\ 2\sqrt{15}$
By the Pythagorean Theorem, $(2x)^2+y^2=BE^2$, and $x^2+(2y)^2=AD^2$.
Plugging in, $4x^2+y^2=16$, and $x^2+4y^2=49$.
Adding the equations, $5x^2+5y^2=65$, and dividing by $5$, $x^2+y^2=13$.

Note that the length $AB^2$ is equal to $(2x)^2+(2y)^2=4x^2+4y^2=4(x^2+y^2)$.
Therefore, the answer is $\sqrt{4(13)}=2\sqrt{13}$ $\fbox{D}$