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Difference between revisions of "2013 AMC 10A Problems/Problem 3"
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==Solution==
 
==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=\boxed{\textbf{(E) }8}</math>
  
  

Revision as of 14:21, 8 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 $BE=\boxed{\textbf{(E) }8}$


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
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