Difference between revisions of "2005 AMC 10A Problems/Problem 23"

m
Line 1: Line 1:
 
==Problem==
 
==Problem==
 
<math>BCDE</math> is a [[square (geometry) | square]]. [[Point]] <math>A</math> is chosen outside of <math>BCDE</math> such that [[angle]] <math>BAC= 120^\circ</math> and <math>AB=AC</math>. Point <math>F</math> is chosen inside <math>BCDE</math> such that the [[triangle]]s <math>ABC</math> and <math>FCD</math> are [[congruent (geometry) | congruent]]. If <math>AF=20</math>, compute the [[area]] of <math>BCDE</math>.  
 
<math>BCDE</math> is a [[square (geometry) | square]]. [[Point]] <math>A</math> is chosen outside of <math>BCDE</math> such that [[angle]] <math>BAC= 120^\circ</math> and <math>AB=AC</math>. Point <math>F</math> is chosen inside <math>BCDE</math> such that the [[triangle]]s <math>ABC</math> and <math>FCD</math> are [[congruent (geometry) | congruent]]. If <math>AF=20</math>, compute the [[area]] of <math>BCDE</math>.  
 
(A)1/6 (B) 1/4 (C) 1/3 (D) 1/2 (E)2/3
 
  
 
<math> \mathrm{(A) \ } \frac{1}{6}\qquad \mathrm{(B) \ } \frac{1}{4}\qquad \mathrm{(C) \ } \frac{1}{3}\qquad \mathrm{(D) \ } \frac{1}{2}\qquad \mathrm{(E) \ } \frac{2}{3} </math>
 
<math> \mathrm{(A) \ } \frac{1}{6}\qquad \mathrm{(B) \ } \frac{1}{4}\qquad \mathrm{(C) \ } \frac{1}{3}\qquad \mathrm{(D) \ } \frac{1}{2}\qquad \mathrm{(E) \ } \frac{2}{3} </math>

Revision as of 13:43, 14 February 2007

Problem

$BCDE$ is a square. Point $A$ is chosen outside of $BCDE$ such that angle $BAC= 120^\circ$ and $AB=AC$. Point $F$ is chosen inside $BCDE$ such that the triangles $ABC$ and $FCD$ are congruent. If $AF=20$, compute the area of $BCDE$.

$\mathrm{(A) \ } \frac{1}{6}\qquad \mathrm{(B) \ } \frac{1}{4}\qquad \mathrm{(C) \ } \frac{1}{3}\qquad \mathrm{(D) \ } \frac{1}{2}\qquad \mathrm{(E) \ } \frac{2}{3}$

Solution

This problem needs a solution. If you have a solution for it, please help us out by adding it.

See also

Invalid username
Login to AoPS