Difference between revisions of "2013 AMC 10A Problems/Problem 3"
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==Solution== | ==Solution== | ||
| − | We know that the area of <math>\triangle | + | We know that the area of <math>\triangle ABE</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> |
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==See Also== | ==See Also== | ||
Revision as of 22:44, 20 November 2013
Problem
Square 
has side length 
. Point 
is on 
, and the area of 
is 
. What is 
?
![[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]](http://latex.artofproblemsolving.com/d/9/a/d9ad7d16a15397560a5daa045621239464589d6e.png)
Solution
We know that the area of is equal to 
. Plugging in 
, we get 
. Dividing, we find that 
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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
