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

Revision as of 00:24, 5 July 2013

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

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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

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

Revision as of 00:24, 5 July 2013

Problem

Solution

See Also