Difference between revisions of "2008 AMC 12B Problems/Problem 13"

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==Problem==
 
==Problem==
  
Vertex <math>E</math> of equilateral <math>\triangle{ABC}</math> is in the interior of unit square <math>ABCD</math>. Let <math>R</math> be the region consisting of all points inside <math>ABCD</math> and outside <math>\triangle{ABC}</math> whose distance from <math>AD</math> is between <math>\frac{1}{3}</math> and <math>\frac{2}{3}</math>. What is the area of <math>R</math>?  
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Vertex <math>E</math> of equilateral <math>\triangle{ABE}</math> is in the interior of unit square <math>ABCD</math>. Let <math>R</math> be the region consisting of all points inside <math>ABCD</math> and outside <math>\triangle{ABC}</math> whose distance from <math>AD</math> is between <math>\frac{1}{3}</math> and <math>\frac{2}{3}</math>. What is the area of <math>R</math>?  
  
 
<math>\textbf{(A)}\ \frac{12-5\sqrt3}{72} \qquad
 
<math>\textbf{(A)}\ \frac{12-5\sqrt3}{72} \qquad

Revision as of 16:41, 31 May 2011

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Problem

Vertex $E$ of equilateral $\triangle{ABE}$ is in the interior of unit square $ABCD$. Let $R$ be the region consisting of all points inside $ABCD$ and outside $\triangle{ABC}$ whose distance from $AD$ is between $\frac{1}{3}$ and $\frac{2}{3}$. What is the area of $R$?

$\textbf{(A)}\ \frac{12-5\sqrt3}{72} \qquad \textbf{(B)}\ \frac{12-5\sqrt3}{36} \qquad \textbf{(C)}\ \frac{\sqrt3}{18} \qquad \textbf{(D)}\ \frac{3-\sqrt3}{9} \qquad \textbf{(E)}\ \frac{\sqrt3}{12}$

See Also

2008 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 12
Followed by
Problem 14
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All AMC 12 Problems and Solutions