2013 AMC 10A Problems/Problem 3

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] size(150); pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); draw((0,0)--(0,10)--(10,10)--(10,0)--(0,0)--(6,10)); dot((0,0)); dot((0,10)); dot((10,10)); dot((10,0)); dot((6,10)); label(" $A$ ",(0,0),SW); label(" $B$ ",(0,10),NW); label(" $C$ ",(10,10),NE); label(" $D$ ",(10,0),SE); label(" $E$ ",(6,10),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 $BE=8$ , $\textbf{(E)}$

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2013 AMC 10A Problems/Problem 3

Problem

Solution