# 2007 AMC 8 Problems/Problem 8

## Problem

In trapezoid $ABCD$, $\overline{AD}$ is perpendicular to $\overline{DC}$, $AD = AB = 3$, and $DC = 6$. In addition, $E$ is on $\overline{DC}$, and $\overline{BE}$ is parallel to $\overline{AD}$. Find the area of $\triangle BEC$. $[asy] defaultpen(linewidth(0.7)); pair A=(0,3), B=(3,3), C=(6,0), D=origin, E=(3,0); draw(E--B--C--D--A--B); draw(rightanglemark(A, D, C)); label("A", A, NW); label("B", B, NW); label("C", C, SE); label("D", D, SW); label("E", E, NW); label("3", A--D, W); label("3", A--B, N); label("6", E, S); [/asy]$

$\textbf{(A)}\ 3 \qquad \textbf{(B)}\ 4.5 \qquad \textbf{(C)}\ 6 \qquad \textbf{(D)}\ 9 \qquad \textbf{(E)}\ 18$

## Solution 1 (Area Formula for Triangles)

Clearly, $ABED$ is a square with side-length $3.$ By segment subtraction, we have $EC = DC - DE = 6 - 3 = 3.$

The area of $\triangle BEC$ is $$\frac12\cdot EC\cdot BE = \frac12\cdot3\cdot3 = \boxed{\textbf{(B)}\ 4.5}.$$ ~Aplus95 (Solution)

~MRENTHUSIASM (Revision)

## Solution 2 (Area Subtraction)

Clearly, $ABED$ is a square with side-length $3.$

Let the brackets denote areas. We apply area subtraction to find the area of $\triangle BEC:$ \begin{align*} [BEC]&=[ABCD]-[ABED] \\ &=\frac{AB+CD}{2}\cdot AD - AB^2 \\ &=\frac{3+6}{2}\cdot 3 - 3^2 \\ &=\boxed{\textbf{(B)}\ 4.5}. \end{align*} ~MRENTHUSIASM