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 3: Line 3:
 
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>?
 +
<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]
+
</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("<math>A</math>",(0,0),SW);
 
−
label("<math>B</math>",(0,10),NW);
 
−
label("<math>C</math>",(10,10),NE);
 
−
label("<math>D</math>",(10,0),SE);
 
−
label("<math>E</math>",(6,10),N);[/asy]
 
  
  
Line 25: Line 35:
  
 
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>
 +
 +
 +
==See Also==
 +
 +
{{AMC10 box|year=2013|ab=A|num-b=2|num-a=4}}
 +
{{AMC12 box|year=2013|ab=A|before=First Problem|num-a=2}}

Revision as of 21:45, 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] 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]


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

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


See Also

2013 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 2 		Followed by
Problem 4
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 10 Problems and Solutions
2013 AMC 12A (Problems • Answer Key • Resources)
Preceded by
First Problem 		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
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_10A_Problems/Problem_3&oldid=50875"