Difference between revisions of "2014 AMC 10A Problems/Problem 22"

(Created page with "==Problem== In rectangle <math>ABCD</math>, <math>AB=20</math> and <math>BC=10</math>. Let <math>E</math> be a point on <math>\overline{CD}</math> such that <math>\angle CBE=15^...")
 
Line 4: Line 4:
  
 
<math> \textbf{(A)}\ \dfrac{20\sqrt3}3\qquad\textbf{(B)}\ 10\sqrt3\qquad\textbf{(C)}\ 18\qquad\textbf{(D)}\ 11\sqrt3\qquad\textbf{(E)}\ 20 </math>
 
<math> \textbf{(A)}\ \dfrac{20\sqrt3}3\qquad\textbf{(B)}\ 10\sqrt3\qquad\textbf{(C)}\ 18\qquad\textbf{(D)}\ 11\sqrt3\qquad\textbf{(E)}\ 20 </math>
 +
 +
==Solution==
 +
 +
==See Also==
 +
 +
{{AMC10 box|year=2014|ab=A|num-b=21|num-a=23}}
 +
{{MAA Notice}}

Revision as of 23:20, 6 February 2014

Problem

In rectangle $ABCD$, $AB=20$ and $BC=10$. Let $E$ be a point on $\overline{CD}$ such that $\angle CBE=15^\circ$. What is $AE$?

$\textbf{(A)}\ \dfrac{20\sqrt3}3\qquad\textbf{(B)}\ 10\sqrt3\qquad\textbf{(C)}\ 18\qquad\textbf{(D)}\ 11\sqrt3\qquad\textbf{(E)}\ 20$

Solution

See Also

2014 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 21
Followed by
Problem 23
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png