69. As shown in the figure, it is known that the graph of the inverse proportion function $y=\frac{k}{x}(k>0)$ passes through the midpoint $\mathrm{D}$ of the hypotenuse $\mathrm{OB}$ of $\mathrm{Rt} O A B$, and intersects the right-angle side $\mathrm{AB}$ at point $\mathrm{C}$. If the area of $O B C$ is 3, then $k=$ $\qquad$
Answer: 2
Solution: Draw a perpendicular line from point $\mathrm{D}$ to the $x$-axis, intersecting the $x$-axis at point $\mathrm{E}$.
By the property of the inverse proportion function, we know that the area of $\triangle ODE$ is equal to the area of $\triangle OAC$.
Therefore, the area of $\triangle OBC$ is equal to the area of quadrilateral DEAB, which is 3.
Let the $x$-coordinate of point $\mathrm{D}$ be $x$, then its $y$-coordinate is $\frac{k}{x}$,
Since $\mathrm{D}$ is the midpoint of $\mathrm{OB}$.
Thus,
$$
EA=x, \quad AB=\frac{2 k}{x}
$$

The area of quadrilateral DEAB can be expressed as
$$
\frac{1}{2}\left(\frac{k}{x}+\frac{2 k}{x}\right) x=3
$$

Solving for $k$, we get
$$
k=2 \text{. }
$$