3. In $\triangle A B C$, let $\angle C=90^{\circ}, \angle A=$ 22. $5^{\circ}, A B=4$. Then the area of $\triangle A B C$ is ( ).
(A) 2
(B) 3
(C) $2 \sqrt{2}$
(D) $2 \sqrt{3}$
3. C.

As shown in Figure 4, draw the median $C M$ on the hypotenuse $A B$. Then
$$
\begin{array}{l}
C M=B M=2, \\
\angle B M C=2 \angle A \\
=45^{\circ} .
\end{array}
$$

Draw $C H \perp A B$ at point $H$. Then $C H=\sqrt{2}$. Therefore, $S_{\triangle A B C}=\frac{1}{2} C H \cdot A B^{\prime}=2 \sqrt{2}$.