5. As shown in the figure, the graphs of the direct proportion functions $y=x$ and $y=a x(a>0)$ intersect with the graph of the inverse proportion function $y=\frac{k}{x}(k>0)$ at points $A$ and $C$, respectively. If the areas of Rt $\triangle A O B$ and Rt $\triangle C O D$ are $S_{1}$ and $S_{2}$, respectively, then the relationship between $S_{1}$ and $S_{2}$ is
(A) $S_{1}>S_{2}$;
(B) $S_{1}=S_{2}$;
(C) $S_{1}<S_{2}$;
(D) Uncertain.
5. (B)

Solution Let the coordinates of point $A$ be $\left(x_{1}, y_{1}\right)$, and the coordinates of point $C$ be $\left(x_{2}, y_{2}\right)$, then
$$
\begin{array}{l}
x y_{1}=x_{2} y_{2}=k . \\
\begin{aligned}
\therefore \quad S_{1} & =\frac{1}{2} O B \cdot A B=\frac{1}{2} x_{1} y_{1}=\frac{1}{2} x_{2} y_{2}=\frac{1}{2} O D \cdot C D \\
& =S_{2} .
\end{aligned}
\end{array}
$$