22. As shown in Figure 2, in rectangle $D E F A$, $A D=3, A F=$ $4, D C=C B=B A$. Then the area of the shaded part is ( ).
(A) 2
(B) $\frac{5}{2}$
(C) 3
(D) $\frac{7}{2}$
(E) 5
22. C.

Let $B E$ intersect $C F$ at point $O$.
In rectangle $D E F A$, we have
$$
\begin{array}{l}
A D / / E F \Rightarrow \triangle B C O \backsim \triangle E F O \\
\Rightarrow \frac{B O}{O E}=\frac{B C}{E F}=\frac{1}{3} \\
\Rightarrow S_{\triangle B C E}=\frac{1}{2} \times 1 \times 4=2, \\
S_{\triangle C O E}=\frac{3}{4} S_{\triangle B C E}=\frac{3}{2} .
\end{array}
$$

Thus, $S_{\text {shaded }}=2 S_{\triangle C O E}=3$.