68. As shown in the figure, a line passing through the origin intersects the graph of the inverse proportion function $y=\frac{k}{x}$ at points $\mathrm{A}$ and $\mathrm{C}$. Perpendiculars are drawn from points $\mathrm{A}$ and $\mathrm{C}$ to the $\mathrm{x}$-axis, with feet at $\mathrm{B}$ and $\mathrm{D}$, respectively. Given that the area of quadrilateral $\mathrm{ABCD}$ is 10, then $k=$ $\qquad$.
Answer: -5
Solution: Since $\mathrm{AB} \perp \mathrm{BD}, \mathrm{CD} \perp \mathrm{BD}$, by the property of the inverse proportion function, it is known that $\mathrm{A}$ and $\mathrm{C}, \mathrm{B}$ and $\mathrm{D}$ are centrally symmetric about point $\mathrm{O}$, so $\mathrm{AB} / / \mathrm{BC}$, and $\mathrm{AB}=\mathrm{BC}$, thus quadrilateral $\mathrm{ABCD}$ is a parallelogram,

Therefore,
$$
\mathrm{S}_{\triangle \mathrm{AOB}}=\mathrm{S}_{\triangle \mathrm{BOC}}=\mathrm{S}_{\triangle \mathrm{COD}}=\mathrm{S}_{\triangle \mathrm{AOD}},
$$

It is easy to know
$$
\mathrm{S}_{\triangle \mathrm{AOB}}=\mathrm{S}_{\triangle \mathrm{COD}}=\frac{1}{2} B O \cdot A B=\frac{k}{2} \text {, }
$$

So
$$
S_{A B C D}=4 S_{A O B}=2|k|=10,
$$

From the graph, we can see that the graph of the inverse proportion function passes through the second and fourth quadrants, so $k<0$,
Therefore, we get
$$
k=-5 \text { . }
$$