# Difference between revisions of "1953 AHSME Problems/Problem 13"

## Problem

A triangle and a trapezoid are equal in area. They also have the same altitude. If the base of the triangle is 18 inches, the median of the trapezoid is:

$\textbf{(A)}\ 36\text{ inches} \qquad \textbf{(B)}\ 9\text{ inches} \qquad \textbf{(C)}\ 18\text{ inches}\\ \textbf{(D)}\ \text{not obtainable from these data}\qquad \textbf{(E)}\ \text{none of these}$

## Solution

We know that the area of a trapezoid is $mh$ where $m$ is the median and the area of a triangle is $\frac{bh}{2}$. We know that $b=18$ and the height of the trapezoid is congruent to the height of the triangle. Thus, we get $\frac{18h}{2}=9h=mh$, so $m=\textbf{(B) }9\text{ inches}$.