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)

2007 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 7	Followed by Problem 9
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All AJHSME/AMC 8 Problems and Solutions

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2007 AMC 8 Problems/Problem 8

Problem

Solution 1 (Area Formula for Triangles)

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