Point $F$ is taken in side $AD$ of square $ABCD$. At $C$ a perpendicular is drawn to $CF$, meeting $AB$ extended at $E$. 
The area of $ABCD$ is $256$ square inches and the area of $\triangle CEF$ is $200$ square inches. Then the number of inches in $BE$ is:

$\textbf{(A)}\ 12 \qquad \textbf{(B)}\ 14 \qquad \textbf{(C)}\ 15 \qquad \textbf{(D)}\ 16 \qquad \textbf{(E)}\ 20$
Because $ABCD$ is a [square](https://artofproblemsolving.com/wiki/index.php/Square), $DC = CB = 16$, $DC \perp DA$, and $CB \perp BA$.  Also, because $\angle DCF + \angle FCB = \angle FCB + \angle BCE = 90^\circ$, $\angle DCF = \angle BCE$.  Thus, by ASA Congruency, $\triangle DCF \cong \triangle BCF$.
From the congruency, $CF = CE$.  Using the area formula for a triangle, $CE = 20$.  Finally, by the [Pythagorean Theorem](https://artofproblemsolving.com/wiki/index.php/Pythagorean_Theorem), $BE = \sqrt{20^2 - 16^2} = 12$, which is answer choice $\boxed{\textbf{(A)}}$.