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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>
| + | #redirect [[2013 AMC 12A Problems/Problem 1]] |
| − | is <math>40</math>. What is <math> BE </math>?
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| − | [asy] | |
| − | size(150);
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| − | pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps);
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| − | draw((0,0)--(0,10)--(10,10)--(10,0)--(0,0)--(6,10));
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| − | dot((0,0));
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| − | dot((0,10));
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| − | dot((10,10));
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| − | dot((10,0));
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| − | dot((6,10));
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| − | label("<math>A</math>",(0,0),SW);
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| − | label("<math>B</math>",(0,10),NW);
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| − | label("<math>C</math>",(10,10),NE);
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| − | label("<math>D</math>",(10,0),SE);
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| − | label("<math>E</math>",(6,10),N);[/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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| − | We know that the area of <math>\triangle ABC</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=8</math>, <math>\textbf{(E)}</math>
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