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2007 AMC 8 Problems/Problem 8: Difference between revisions

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<math>\triangle BEC</math>.
<math>\triangle BEC</math>.


<center>[[Image:AMC8_2007_8.png]]</center>
<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>


<math>\text{(A)}\ 3 \qquad \text{(B)}\ 4.5 \qquad \text{(C)}\ 6 \qquad \text{(D)}\ 9 \qquad \text{(E)}\ 18</math>
<math>\text{(A)}\ 3 \qquad \text{(B)}\ 4.5 \qquad \text{(C)}\ 6 \qquad \text{(D)}\ 9 \qquad \text{(E)}\ 18</math>


== Solution ==
== Solution ==

Revision as of 12:14, 9 December 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 $\triangle BEC$.

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

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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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