Difference between revisions of "2013 AMC 10A Problems/Problem 3"

Revision as of 11:06, 4 July 2013

Problem

Square $ABCD$ has side length $10$ . Point $E$ is on $\overline{BC}$ , and the area of $\triangle ABE$ is $40$ . What is $BE$ ? $[asy] pair A,B,C,D,E; A=(0,0); B=(0,50); C=(50,50); D=(50,0); E = (30,50); draw(A--B); draw(B--E); draw(E--C); draw(C--D); draw(D--A); draw(A--E); dot(A); dot(B); dot(C); dot(D); dot(E); label("A",A,SW); label("B",B,NW); label("C",C,NE); label("D",D,SE); label("E",E,N); [/asy]$

$\textbf{(A)}\ 4\qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 6\qquad\textbf{(D)}\ 7\qquad\textbf{(E)}\ 8$

Solution

We know that the area of $\triangle ABC$ is equal to $\frac{AB(BE)}{2}$ . Plugging in $AB=10$ , we get $80 = 10BE$ . Dividing, we find that $BE=\boxed{\textbf{(E) }8}$

2013 AMC 10A (Problems • Answer Key • Resources)
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2013 AMC 12A (Problems • Answer Key • Resources)
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The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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 {{AMC10 box|year=2013|ab=A|num-b=2|num-a=4}}
 {{AMC12 box|year=2013|ab=A|before=First Problem|num-a=2}}
+{{MAA Notice}}

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Difference between revisions of "2013 AMC 10A Problems/Problem 3"

Revision as of 11:06, 4 July 2013

Problem

Solution

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