2005 AMC 10A Problems/Problem 23: Difference between revisions
No edit summary |
No edit summary |
||
| Line 1: | Line 1: | ||
BCDE is a square. Point A is chosen outside of BCDE such that angle BAC= 120 and AB=AC. Point F is chosen inside BCDE such that the | ==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>. | |||
(A)1/6 (B) 1/4 (C) 1/3 (D) 1/2 (E)2/3 | (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> | |||
==Solution== | |||
{{solution}} | |||
==See also== | |||
[[Category:Introductory Geometry Problems]] | |||
Revision as of 12:43, 14 February 2007
Problem
is a square. Point 
is chosen outside of such that angle 
and 
. Point 
is chosen inside such that the triangles 
and 
are congruent. If 
, compute the area of .
(A)1/6 (B) 1/4 (C) 1/3 (D) 1/2 (E)2/3
Solution
This problem needs a solution. If you have a solution for it, please help us out by adding it.
