3. In pentagon $A B C D E$, $\angle A=\angle C=90^{\circ}$, $A B=B C=D E=A E+C D=3$. Then the area of this pentagon is $(\quad)$.
(A) 9
(B) 10.5
(C) 12
(D) 13.5
3. A.

As shown in Figure 3, extend $DC$ to point $F$ such that $CF = AE$, and connect $BE$, $BD$, and $BF$. Then,
$$
DF = DE = 3.
$$

Also, $\triangle BCF \cong \triangle BAE$, so
$$
BE = BF.
$$

Since $BD = BD$, we have
$$
\triangle BED \cong \triangle BFD.
$$

Therefore, $S_{\triangle BRD} = S_{\triangle BFD}$. Hence, $S_{\text{pentagon } 1BCDE} = S_{\text{quadrilateral } BRDF} = 2S_{\triangle BFD}$
$$
= 2 \times \frac{1}{2} DF \cdot BC = 9.
$$