Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2013 AMC 10A Problems/Problem 3"
Line 1: Line 1:
 +
==Problem==
 +
 
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>
 
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>
 
is <math>40</math>. What is <math> BE </math>?
 
is <math>40</math>. What is <math> BE </math>?
Line 19: Line 21:
  
 
<math> \textbf{(A)}\ 4\qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 6\qquad\textbf{(D)}\ 7\qquad\textbf{(E)}\ 8 </math>
 
<math> \textbf{(A)}\ 4\qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 6\qquad\textbf{(D)}\ 7\qquad\textbf{(E)}\ 8 </math>
 +
 +
==Solution==
  
 
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>
 
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>

Revision as of 18:27, 7 February 2013

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 $BE=8$, $\textbf{(E)}$

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_10A_Problems/Problem_3&oldid=50840"