# 1965 AHSME Problems/Problem 8

## Problem

One side of a given triangle is 18 inches. Inside the triangle a line segment is drawn parallel to this side forming a trapezoid whose area is one-third of that of the triangle. The length of this segment, in inches, is:

$\textbf{(A)}\ 6\sqrt {6} \qquad \textbf{(B) }\ 9\sqrt {2} \qquad \textbf{(C) }\ 12 \qquad \textbf{(D) }\ 6\sqrt{3}\qquad \textbf{(E) }\ 9$

## Solution

$[asy] draw((0,0)--(18,0)); dot((0,0)); label("A", (-1.5,-0.5)); dot((18,0)); label("B",(19.5,-0.5)); label("18", (9,-1.5)); draw((0,0)--(5,10)--(18,0)); dot((5,10)); label("E", (6,11.5)); draw((0.918,1.835)--(15.615,1.835)); dot((0.918,1.835)); label("D", (-0.65,2.25)); dot((15.615,1.835)); label("C", (17.25,2.25)); [/asy]$

Let the given triangle be $\triangle ABE$ with $AB=18$. Also, let $\overline{CD} \parallel \overline{AB}$ with $C$ on $\overline{BE}$ and $D$ on $\overline{AE}$. We know from the problem that $[ABCD]=\frac{1}{3}[\triangle ABE]$, so $[\triangle DCE]=\frac{2}{3}[\triangle ABE]$. Because $\overline{DC} \parallel \overline{AB}$, $\triangle DCE \sim \triangle ABE$ by AA similarity. Because the ratio of these two triangles' areas is $\frac{2}{3}$, the ratio of their sides must be $\sqrt{\frac{2}{3}}=\frac{\sqrt{6}}{3}$. Thus, $CD=\frac{\sqrt{6}}{3}AB=\frac{\sqrt{6}}{3}*18=\boxed{6\sqrt{6}}$, which is answer choice $\fbox{A}$.