11. Given that $A$ and $B$ are two moving points on the ellipse $\frac{x^{2}}{9}+\frac{y^{2}}{4}=1$, $O$ is the origin, and $O A \perp O B$. Then the minimum value of $S_{\triangle A O B}$ is
11. $\frac{36}{13}$.

Let $A\left(r_{1} \cos \theta, r_{1} \sin \theta\right), B\left(-r_{2} \sin \theta, r_{2} \cos \theta\right)$. Substituting the coordinates of points $A$ and $B$ into the ellipse equation, we get
$$
\begin{array}{l}
\frac{r_{1}^{2} \cos ^{2} \theta}{9}+\frac{r_{1}^{2} \sin ^{2} \theta}{4}=1, \\
\frac{r_{2}^{2} \sin ^{2} \theta}{9}+\frac{r_{2}^{2} \cos ^{2} \theta}{4}=1 .
\end{array}
$$

Then $r_{1}^{2}=\frac{36}{4 \cos ^{2} \theta+9 \sin ^{2} \theta}$.
$$
r_{2}^{2}=\frac{36}{4 \sin ^{2} \theta+9 \cos ^{2} \theta} \text {. }
$$

Since $O A \perp O B$, we have
$$
\begin{array}{l}
S_{\triangle A O B}=\frac{1}{2} r_{1} r_{2} \\
=\frac{18}{\sqrt{\left(4 \cos ^{2} \theta+9 \sin ^{2} \theta\right)\left(4 \sin ^{2} \theta+9 \cos ^{2} \theta\right)}} \\
=\frac{18}{\sqrt{97 \sin ^{2} \theta \cdot \cos ^{2} \theta+36\left(\sin ^{4} \theta+\cos ^{4} \theta\right)}} \\
=\frac{18}{\sqrt{36+25 \sin ^{2} \theta \cdot \cos ^{2} \theta}}=\frac{18}{\sqrt{36+\frac{25}{4} \sin ^{2} 2 \theta}} \\
\geqslant \frac{18}{\sqrt{36+\frac{25}{4}}}=\frac{36}{13} .
\end{array}
$$

When and only when $|\sin 2 \theta|=1$, i.e., $\theta=k \pi \pm \frac{\pi}{4}(k \in \mathbf{Z})$, $S_{\triangle A O B}$ reaches its minimum value $\frac{36}{13}$.