3. As shown in Figure 2, in trapezoid $A B C D$, $A B / / D C, A B$ $\perp B C, E$ is the midpoint of $A D$, $A B+B C+C D=6, B E$ $=\sqrt{5}$. Then the area of trapezoid $A B C D$ is ( ).
(A) 13
(B) 8
(C) $\frac{13}{2}$
(D) 4
3. D.

As shown in Figure 10, draw $E F / / A B$ intersecting $B C$ at point $F$. Then
$$
\begin{array}{c}
B F=\frac{1}{2} B C, \\
E F=\frac{1}{2}(A B+C D)=\frac{1}{2}(6-B C) .
\end{array}
$$

Since $A B \perp B C$, it follows that $E F \perp B C$.
Therefore, in the right triangle $\triangle B F E$, we have
$$
\begin{array}{l}
E F^{2}+B F^{2}=B E^{2} \\
\Rightarrow\left[\frac{1}{2}(6-B C)\right]^{2}+\left(\frac{1}{2} B C\right)^{2}=(\sqrt{5})^{2} \\
\Rightarrow B C^{2}-6 B C+8=0 \\
\Rightarrow B C=2 \text { or } 4 \\
\Rightarrow E F=2 \text { or } 1 \\
\Rightarrow S_{\text {trapezoid } A B C D}=E F \cdot B C=4 .
\end{array}
$$