Problem:
In the figure, ABCD is a 2×2 square, E is the midpoint of AD, and F is on BE. If CF is perpendicular to BE, then the area of quadrilateral CDEF is
Answer Choices:
A. 2
B. 3−23
C. 511
D. 5
E. 49
Solution:
In right triangle BAE,BE=22+12=5. Since △CFB∼△BAE, it follows that [CFB]=(BECB)2⋅[BAE]=(52)2⋅21(2⋅1)=54. Then [CDEF]=[ABCD]−[BAE]−[CFB]=4−1−54=511.
OR
Draw the figure in the plane as shown with B at the origin. An equation of the line BE is y=2x, and, since the lines are perpendicular, an equation of the line CF is y=−21(x−2). Solve these two equations simultaneously to get F=(2/5,4/5) and
