Difference between revisions of "2014 AMC 10A Problems/Problem 22"
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==Solution== | ==Solution== | ||
− | Note that <math>\tan 15^\circ=\frac{EC}{10} \Rightarrow EC=20-10 \sqrt 3</math>. Therefore, we have <math> | + | Note that <math>\tan 15^\circ=\frac{EC}{10} \Rightarrow EC=20-10 \sqrt 3</math>. Therefore, we have <math>DE=10\sqrt 3</math>. Since <math>ADE</math> is a <math>30-60-90</math> triangle, <math>AE=2 \cdot DE=2 \cdot 10=\boxed{\textbf{(E)} \: 20}</math> |
==See Also== | ==See Also== |
Revision as of 15:36, 18 February 2014
Problem
In rectangle , and . Let be a point on such that . What is ?
Solution
Note that . Therefore, we have . Since is a triangle,
See Also
2014 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 21 |
Followed by Problem 23 | |
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All AMC 10 Problems and Solutions |
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