18. B.

From the given information,
$$
\begin{array}{l}
B D=\sqrt{B C^{2}+C D^{2}}=5 \\
\Rightarrow B D^{2}+A B^{2}=A D^{2} \\
\Rightarrow \angle A B D=90^{\circ} .
\end{array}
$$

Therefore, $S=S_{\triangle A B D}-S_{\triangle B C D}$
$$
=\frac{1}{2} \times 5 \times 12-\frac{1}{2} \times 3 \times 4=24 \text {. }
$$
18. As shown in Figure 3, in the concave quadrilateral $A B C D$, $\angle B C D$ $=90^{\circ}, A B=12, B C=4, C D=3, A D=13$. Then the area $S$ of the concave quadrilateral $A B C D$ is ( ).
(A) 12
(B) 24
(C) 26
(D) 30
(E) 36