In the figure, $ABCD$ is a $2 \times 2$ square, $E$ is the midpoint of $\overline{AD}$, and $F$ is on $\overline{BE}$.  If $\overline{CF}$ is perpendicular to $\overline{BE}$, then the area of quadrilateral $CDEF$ is

$\textbf{(A)}\ 2\qquad\textbf{(B)}\ 3-\frac{\sqrt{3}}{2}\qquad\textbf{(C)}\ \frac{11}{5}\qquad\textbf{(D)}\ \sqrt{5}\qquad\textbf{(E)}\ \frac{9}{4}$
Since $\angle EBA = \angle FCB$ and $\angle FBC = \angle AEB$, we have $\triangle ABE \sim \triangle FCB$.
$\frac{AB}{FC} = \frac{BE}{CB} = \frac{EA}{BF}$
$\frac{2}{FC} = \frac{\sqrt{5}}{2} = \frac{1}{BF}$
From those two equations, we find that $CF = \frac{4}{\sqrt{5}}$ and $BF = \frac{2}{\sqrt{5}}$
Now that we have $BF$ and $CF$, we can find the area of the bottom triangle $\triangle CFB$:  $\frac{1}{2}\cdot \frac{4}{\sqrt{5}} \cdot \frac{2}{\sqrt{5}} = \frac{4}{5}$
The area of left triangle $\triangle BEA$ is $\frac{1}{2}\cdot 2 \cdot 1 = 1$
The area of the square is $4$.
Thus, the area of the remaining quadrilateral is $4 - 1 - \frac{4}{5} = \frac{11}{5}$, and the answer is $\boxed{C}$