2013 AMC 12A Problems/Problem 1

Problem

Square $ABCD$ has side length $10$ . Point $E$ is on $\overline{BC}$ , and the area of $\bigtriangleup ABE$ is $40$ . What is $BE$ ?

$[asy] pair A,B,C,D,E; A=(0,0); B=(0,50); C=(50,50); D=(50,0); E = (40,50); draw(A--B); draw(B--E); draw(E--C); draw(C--D); draw(D--A); draw(A--E); dot(A); dot(B); dot(C); dot(D); dot(E); label("A",A,SW); label("B",B,NW); label("C",C,NE); label("D",D,SE); label("E",E,N); [/asy]$

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

Solution

We are given that the area of $\triangle ABE$ is $40$ , and that $AB = 10$ .

The area of a triangle is:

$A = \frac{bh}{2}$

Using $AB$ as the height of $\triangle ABE$ ,

$40 = \frac{10b}{2}$

and solving for $b$ ,

$b = 8$ , which is $\boxed{\textbf{(E)}}$

Video Solution (CREATIVE THINKING)

https://youtu.be/wW_GidbHnnM

~Education, the Study of Everything

Video Solution

https://www.youtube.com/watch?v=2vf843cvVzo?t=0 ~sugar_rush

2013 AMC 10A (Problems • Answer Key • Resources)
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2013 AMC 12A (Problems • Answer Key • Resources)
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All AMC 12 Problems and Solutions

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

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2013 AMC 12A Problems/Problem 1

Contents

Problem

Solution

Video Solution (CREATIVE THINKING)

Video Solution

See also