10. (20 points) Given a line $l$ intersects the ellipse $C: \frac{x^{2}}{2}+y^{2}=1$ at points $A, B$, and $O$ is the origin.
(1) Find the maximum value of the area of $\triangle A O B$, and the equation of the line $l$ at this time;
(2) Let the set $L=\{l \mid$ line $l$ makes the area of $\triangle A O B$ take the maximum value $\}, l_{1}, l_{2}, l_{3}, l_{4} \in L$ and satisfy $l_{1} / / l_{2}, l_{3} / / l_{4}$, $k_{1}, k_{2}, k_{3}, k_{4}$ are the slopes of $l_{1}, l_{2}, l_{3}, l_{4}$ respectively and satisfy $k_{1}+k_{2}+k_{3}+k_{4}=0$, find the minimum value of the area of the quadrilateral formed by these four lines.
10. (1) When the slope of $l$ does not exist, we can set $l: x=t$, then
$$
S_{\triangle A O B}=\frac{1}{2}|t| \cdot 2 \sqrt{1-\frac{t^{2}}{2}}=\frac{\sqrt{2}}{2} \sqrt{t^{2}\left(2-t^{2}\right)} \leqslant \frac{\sqrt{2}}{2} .
$$

Equality holds if and only if $t^{2}=2-t^{2}$, i.e., $t= \pm 1$.
When the slope of $l$ exists, we can set $l: y=k x+m$, from
$$
\left\{\begin{array}{l}
y=k x+m \\
x^{2}+2 y^{2}=2
\end{array}\right.
$$

we get
$$
\left(1+2 k^{2}\right) x^{2}+4 k m x+2\left(m^{2}-1\right)=0,
$$

so
$$
\Delta=16 k^{2} m^{2}-8\left(1+2 k^{2}\right)\left(m^{2}-1\right)=8\left(1+2 k^{2}-m^{2}\right),
$$

so
$$
\begin{aligned}
S_{\triangle A O B} & =\frac{1}{2}|A B| \cdot h=\frac{1}{2} \sqrt{1+k^{2}} \cdot \frac{\sqrt{8\left(1+2 k^{2}-m^{2}\right)}}{1+2 k^{2}} \cdot \frac{|m|}{\sqrt{1+k^{2}}} \\
& =\sqrt{2} \frac{\sqrt{\left(1+2 k^{2}-m^{2}\right) m^{2}}}{1+2 k^{2}} \\
& \leqslant \sqrt{2} \cdot \frac{\frac{1}{2}\left(1+2 k^{2}\right)}{1+2 k^{2}}=\frac{\sqrt{2}}{2} .
\end{aligned}
$$

In the above, $h$ represents the distance from $O$ to $A B$. Equality holds if and only if $m^{2}=1+2 k^{2}-m^{2}$, i.e., $m^{2}=k^{2}+\frac{1}{2}$.

In summary, the maximum value of $S_{\triangle A O B}$ is $\frac{\sqrt{2}}{2}$, at this time the equation of the line $l$ is $x= \pm 1$ or $y=k x \pm \sqrt{k^{2}+\frac{1}{2}}$ $(k \in \mathbf{R})$.
(2) By symmetry, the lines
$$
\begin{array}{l}
l_{1}: y=k x+\sqrt{k^{2}+\frac{1}{2}}, \\
l_{2}: y=k x-\sqrt{k^{2}+\frac{1}{2}}, \\
l_{3}: y=-k x+\sqrt{k^{2}+\frac{1}{2}}, \\
l_{4}: y=-k x-\sqrt{k^{2}+\frac{1}{2}} .
\end{array}
$$

Here, $k>0$. So the quadrilateral formed by these four lines is a rhombus, then
$$
S=4 \cdot \frac{1}{2} \sqrt{k^{2}+\frac{1}{2}} \cdot \frac{\sqrt{k^{2}+\frac{1}{2}}}{k}=2\left(k+\frac{1}{2 k}\right) \geqslant 2 \sqrt{2},
$$

Equality holds if and only if $k=\frac{1}{2 k}$, i.e., $k=\frac{\sqrt{2}}{2}$. Therefore, the minimum value of the area of the quadrilateral is $2 \sqrt{2}$.