Difference between revisions of "2013 AMC 12A Problems/Problem 1"

(See also)
Line 49: Line 49:
 
== See also ==
 
== See also ==
 
{{AMC12 box|year=2013|ab=A|before=First Question|num-a=2}}
 
{{AMC12 box|year=2013|ab=A|before=First Question|num-a=2}}
 +
 +
[[Category:Introductory Geometry Problems]]

Revision as of 08:54, 7 April 2013

Problem

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

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

[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]

Solution

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

The area of a triangle:

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

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

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

and solving for b,

$b = 8$, which is $E$

See also

2013 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
First Question
Followed by
Problem 2
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 12 Problems and Solutions