3. As shown in Figure 1, let $P\left(x_{p}, y_{p}\right)$ be a point on the graph of the inverse proportion function $y=\frac{2}{x}$ in the first quadrant of the Cartesian coordinate system $x O y$. Draw lines parallel to the $x$-axis and $y$-axis through point $P$, intersecting the graph of $y=\frac{10}{x}$ in the first quadrant at points $A$ and $B$, respectively. Then the area of $\triangle A O B$ is ( ).
(A) 26
(B) 24
(C) 22
(D) 20
3. B.

Connect $O P$.
Then $S_{\triangle A O B}=S_{\triangle A O P}+S_{\triangle P O B}+S_{\triangle A P B}$
$$
\begin{aligned}
= & \frac{1}{2}\left(\frac{10}{y_{p}}-x_{p}\right) y_{p}+\frac{1}{2}\left(\frac{10}{x_{p}}-y_{p}\right) x_{p}+ \\
& \frac{1}{2}\left(\frac{10}{y_{p}}-x_{p}\right)\left(\frac{10}{x_{p}}-y_{p}\right) \\
= & \frac{1}{2} \cdot \frac{100}{x_{p} y_{p}}-\frac{1}{2} x_{p} y_{p} \\
= & 24 .
\end{aligned}
$$