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| − | ==Problem==
| + | #redirect [[2013 AMC 12A Problems/Problem 1]] |
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| − | Square <math> ABCD </math> has side length <math>10</math>. Point <math>E</math> is on <math>\overline{BC}</math>, and the area of <math> \triangle ABE </math>
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| − | is <math>40</math>. What is <math> BE </math>?
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| − | <asy>
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| − | pair A,B,C,D,E;
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| − | A=(0,0);
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| − | B=(0,50);
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| − | C=(50,50);
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| − | D=(50,0);
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| − | E = (30,50);
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| − | draw(A--B);
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| − | draw(B--E);
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| − | draw(E--C);
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| − | draw(C--D);
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| − | draw(D--A);
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| − | draw(A--E);
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| − | dot(A);
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| − | dot(B);
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| − | dot(C);
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| − | dot(D);
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| − | dot(E);
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| − | label("A",A,SW);
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| − | label("B",B,NW);
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| − | label("C",C,NE);
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| − | label("D",D,SE);
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| − | label("E",E,N);
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| − | </asy>
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| − | <math> \textbf{(A)}\ 4\qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 6\qquad\textbf{(D)}\ 7\qquad\textbf{(E)}\ 8 </math>
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| − | ==Solution==
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| − | We know that the area of <math>\triangle ABE</math> is equal to <math>\frac{AB(BE)}{2}</math>. Plugging in <math>AB=10</math>, we get <math>80 = 10BE</math>. Dividing, we find that <math>BE=\boxed{\textbf{(E) }8}</math>
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| − | ==See Also==
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| − | | |
| − | {{AMC10 box|year=2013|ab=A|num-b=2|num-a=4}}
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| − | {{AMC12 box|year=2013|ab=A|before=First Problem|num-a=2}}
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| − | {{MAA Notice}}
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