Difference between revisions of "2013 AMC 10A Problems/Problem 3"
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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> | 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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+ | ==Video Solution== | ||
+ | https://www.youtube.com/watch?v=2vf843cvVzo?t=0 | ||
+ | |||
+ | ~sugar_rush | ||
==See Also== | ==See Also== |
Revision as of 22:18, 23 November 2020
Contents
Problem
Square has side length . Point is on , and the area of is . What is ?
Solution
We know that the area of is equal to . Plugging in , we get . Dividing, we find that
Video Solution
https://www.youtube.com/watch?v=2vf843cvVzo?t=0
~sugar_rush
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.