52. As shown in the figure, in quadrilateral $A B C D$, $A B=7, B C=24, C D=20, D A=15, \angle B=90^{\circ}$, then the area of quadrilateral $A B C D$ is . $\qquad$
Answer: 234.

Because
$$
\angle B=90^{\circ},
$$

By the Pythagorean theorem, we get
$$
A C^{2}=A B^{2}+B C^{2}=7^{2}+24^{2}=49+576=625 \text {, }
$$

So
$$
A C=25 \text {. }
$$

Because
$$
A D^{2}+D C^{2}=15^{2}+20^{2}=225+400=625=A C^{2},
$$

By the converse of the Pythagorean theorem, we get
$$
\angle D=90^{\circ} \text {, }
$$

So
$$
\begin{aligned}
S_{\text {quadrilateral } A B C D} & =S_{\triangle A B C}+S_{\triangle A D C} \\
& =\frac{1}{2} \times 7 \times 24+\frac{1}{2} \times 15 \times 20 \\
& =84+150 \\
& =234
\end{aligned}
$$