# Difference between revisions of "1951 AHSME Problems/Problem 49"

## Problem

The medians of a right triangle which are drawn from the vertices of the acute angles are $5$ and $\sqrt{40}$. The value of the hypotenuse is:

$\textbf{(A)}\ 10\qquad\textbf{(B)}\ 2\sqrt{40}\qquad\textbf{(C)}\ \sqrt{13}\qquad\textbf{(D)}\ 2\sqrt{13}\qquad\textbf{(E)}\ \text{none of these}$

## Solution

We will proceed by coordinate bashing.

Call the first leg $2a$, and the second leg $2b$ (We are using the double of a variable to avoid any fractions)

Notice that we want to find $\sqrt{(2a)^2+(2b)^2}$

Two equations can be written for the two medians: $a^2 + 4b^2 = 40$ and $4a^2+b^2= 25$.

Add them together and we get $5a^2+5b^2=65$,

Dividing by 5 gives $x^2+y^2=13$

Multiplying this by 4 gives $4x^2+4y^2=52\implies (2x)^2+(2y)^2=52$, just what we need to find the hypotenuse. Recall that he hypotenuse is $\sqrt{(2a)^2+(2b)^2}$. The value inside the radical is equal to $52$, so the hypotenuse is equal to $\sqrt{52}=\boxed{\textbf{(D)}\ 2\sqrt{13}}$