32. As shown in the figure, $P$ is a point inside the square $A B C D$, and $\angle A P B=90^{\circ}, A C, B D$ intersect at $O$. It is known that $A P=3 \mathrm{~cm}$, $B P=5 \mathrm{~cm}$. Find the area of triangle $O B P$.
32. As shown in the figure, $P$ is a point inside the square $A B C D$, and $\angle A P B=90^{\circ}, A C, B D$ intersect at $O$. It is known that $A P=3 \mathrm{~cm}$, $B P=5 \mathrm{~cm}$. Find the area of triangle $O B P$.
【Analysis】From the given conditions, we can find $A B, A O, B O$, and the areas of triangles $A O B, A P B$ are $17 \div 2, 15 \div 2$ respectively; then using the proportional relationship and the fact that the ratio of the areas of similar triangles is the square of the similarity ratio, we find that $S \triangle O B E$ is $17 X=\frac{\sqrt{17}}{8}$, and $S \triangle P A E$ is $9 X=9 \times \frac{1}{8}=\frac{9}{8}$; then using the area formula of the triangle to find the lengths of $P E$ and $O E$ as $\frac{3}{4}, \frac{\sqrt{17}}{4}$; then using $\triangle O E H \sim \triangle A E P$, we find $O H=\$ \backslash \text { frac }\{\backslash \text { sqrt }\{17\}\}\{4\} \times 3 \div \text { frac }\{3 \backslash \text { sqrt }\{17\}\}\{4\}=1$; finally, we can find the area of the required triangle.
【Solution】Solution: $\because$ In the square $A B C D$, $A C, B D$ intersect at $O, \angle A P B=90^{\circ}$
$$
\begin{array}{l}
\therefore A O: A B=1:: \sqrt{ } 2, A B=\sqrt{ } 3^{2}+5^{2}, B O=A O=\sqrt{ } 17 \\
\because \triangle O B E \sim \triangle P A E \\
\therefore S \triangle O B E: S \triangle P A E=B O^{2}: A P^{2}=\sqrt{ } 17^{2}: 3^{2}=17: 9
\end{array}
$$

Let $S \triangle O B E$ be $17 X$, then $S \triangle P A E$ is $9 X$
$$
\begin{aligned}
\because S \triangle A E B=S \triangle A O B-S \triangle O B E=S A B E-S \triangle P A E \\
\therefore \sqrt{ } 17 \times \sqrt{ } 17 \div 2-17 X=3 \times 5 \div 2-9 X
\end{aligned}
$$

Solving for $X=\frac{1}{8}, S \triangle O B E=17 X=\frac{\sqrt{17}}{8}, S \triangle A P E=9 X=9 \times \frac{1}{8}=\frac{9}{8}, P E=\frac{9}{8} \times 2 \div 3=\frac{3}{4}$, $O E=\frac{17}{8} \times 2 \div \sqrt{17}=\frac{\sqrt{17}}{4}$
$\because \triangle O E H \sim \triangle A E P$
$$
\begin{array}{l}
\therefore O H=\$ \backslash \text { frac }\{\backslash \text { sqrt }\{17\}\}\{4\} \times 3 \div \text { frac }\{3 \backslash \text { sqrt }\{17\}\}\{4\}=1 \\
\therefore S \triangle B P O=B P \times O H \div 2=5 \times 1 \div 2=2.5
\end{array}
$$

Answer: The area of triangle $O B P$ is 2.5 square centimeters.