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Difference between revisions of "2007 AMC 8 Problems/Problem 8"

(Created page with '== Problem == In trapezoid <math>ABCD</math>, <math>AD</math> is perpendicular to <math>DC</math>, <math>AD</math> = <math>AB</math> = <math>3</math>, and <math>DC</math> = <mat…')
 
(→‎Solution)
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We are trying to find the area of <math>\triangle BEC</math>.
 
We are trying to find the area of <math>\triangle BEC</math>.
  
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So, <math>\frac{1}{2} * 3 * 3 = 4.5</math>, <math>\Boxed{B}</math>
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So, <math>\frac{1}{2} * 3 * 3 = 4.5</math>, <math>\boxed{B}</math>
 +
 
 +
<center>[[Image:AMC8_2007_8S.png]]</center>

Revision as of 17:10, 15 February 2010

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
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