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]=(5​2​)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
