18.23 In the convex quadrilateral $ABCD$, the sides $AB, BC, CD$, and $DA$ are $3, 4, 12$, and $13$, respectively, and $\angle CBA$ is a right angle. The area of the quadrilateral is 
(A) 32.
(B) 36.
(C) 39.
(D) 42.
(E) 48.
(31st American High School Mathematics Examination, 1980)
[Solution] Connect $A C$. In the right triangle $\triangle A B C$, according to the Pythagorean theorem, $A C=\sqrt{3^{2}+4^{2}}=5$. Since $5^{2}+12^{2}=13^{2}$, by the converse of the Pythagorean theorem, $\triangle D A C$ is a right triangle.
$$
\begin{aligned}
\therefore \quad S_{A B C D} & =S_{\triangle A B C}+S_{\triangle A D C} \\
& =\frac{1}{2} \cdot 3 \cdot 4+\frac{1}{2} \cdot 5 \cdot 12 \\
& =36 .
\end{aligned}
$$

Therefore, the answer is $(B)$.