2007 AMC 8 Problems/Problem 8

Problem

In trapezoid $ABCD$, $AD$ is perpendicular to $DC$, $AD$ = $AB$ = $3$, and $DC$ = $6$. In addition, $E$ is on $DC$, and $BE$ is parallel to $AD$. 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]

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

Solution

We know that $ABED$ is a square with side length $3$. We subtract $DC$ and $DE$ to get the length of $EC$.

$EC = DC - DE = 6 - 3 = 3$

We are trying to find the area of $\triangle BEC$.

So, $\frac{1}{2} \cdot 3 \cdot 3 = \boxed{\textbf{(B)}\ 4.5}$

See Also

2007 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 7
Followed by
Problem 9
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AJHSME/AMC 8 Problems and Solutions

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