2. In rectangle $A B C D$, it is known that $A D=5, A B=$ $10, E, F$ are two points outside the rectangle, $B E=D F=4$, $A F=C E=3$. Then $E F=(\quad)$.
(A) $4 \sqrt{15}$
(B) 15
(C) $\sqrt{221}$
(D) $10 \sqrt{2}$
2. C.

As shown in Figure 3, it is easy to see that $\angle A F D=\angle B E C=90^{\circ}$.
Then $\triangle B E C \cong \triangle D F A \Rightarrow \angle D A F=\angle B C E$.
Extend $F A$ and $E B$ to intersect at point $G$.
Since $\angle G A B=90^{\circ}-\angle D A F=\angle A D F$, $\angle G B A=90^{\circ}-\angle C B E=\angle B C E=\angle D A F$,

we know $\triangle B G A \backsim \triangle A F D$, and $\angle A G B=90^{\circ}$.
Thus, $A G=8, B G=6$.
Therefore, $G F=11, G E=10$.
Hence, $E F=\sqrt{G E^{2}+G F^{2}}=\sqrt{221}$.