Difference between revisions of "2005 AMC 10A Problems/Problem 23"
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 13:43, 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.