# 1958 AHSME Problems/Problem 43

## Problem

$\overline{AB}$ 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:

$\textbf{(A)}\ 10\qquad \textbf{(B)}\ 5\sqrt{3}\qquad \textbf{(C)}\ 5\sqrt{2}\qquad \textbf{(D)}\ 2\sqrt{13}\qquad \textbf{(E)}\ 2\sqrt{15}$

## Solution

$[asy] import geometry; unitsize(50); pair C = (0,0), B = (3,0), A = (0, 4); pair AC = midpoint(A--C), BC = midpoint(B--C); draw(A--B--C--A); draw(A--C, StickIntervalMarker(2, 1)); draw(B--C, StickIntervalMarker(2, 2)); draw(B--AC); draw(A--BC); dot(AC); dot(BC); MP("A", A, W); MP("B", B, E); MP("C", C, W); MP("E", AC, W); MP("D", BC, S); label("y", A--AC, W); label("y", AC--C, W); label("x", B--BC, S); label("x", BC--C, S); draw(rightanglemark(A, C, B)); [/asy]$

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 $\sqrt{4(13)}=2\sqrt{13}$ $\fbox{D}$