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==See Also==
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{{AMC8 box|year=2007|num-b=7|num-a=9}}

Revision as of 23:35, 12 November 2012

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 $\triangle BEC$.

AMC8 2007 8.png

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


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, $\frac{1}{2} * 3 * 3 = 4.5$, $\boxed{B}$

AMC8 2007 8S.png

See Also

2007 AMC 8 (Problems • Answer Key • Resources)
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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