11.A. In the Cartesian coordinate system $x O y$, the graph of the linear function $y = kx + b (k \neq 0)$ intersects the positive half-axes of the $x$-axis and $y$-axis at points $A$ and $B$, respectively, and makes the area of $\triangle O A B$ equal to $|O A| + |O B| + 3$.
(1) Express $k$ in terms of $b$;
(2) Find the minimum value of the area of $\triangle O A B$.
Three, 11.A.(1) Let $x=0$, we get $y=b(b>0)$; let $y=0$, we get $x=-\frac{b}{k}>0(k2)$.
(2) From (1), we know
$$
\begin{array}{l} 
S_{\triangle O A B}=\frac{1}{2} b\left(-\frac{b}{k}\right)=\frac{b(b+3)}{b-2} \\
=\frac{(b-2)^{2}+7(b-2)+10}{b-2} \\
=b-2+\frac{10}{b-2}+7 \\
=\left(\sqrt{b-2}-\sqrt{\frac{10}{b-2}}\right)^{2}+7+2 \sqrt{10} \\
\geqslant 7+2 \sqrt{10},
\end{array}
$$

When and only when $b-2=\frac{10}{b-2}$, we have $S_{\triangle O A B}=7+$ $2 \sqrt{10}$, that is, when $b=2+\sqrt{10}, k=-1$, the equality in the inequality holds.
Therefore, the minimum value of the area of $\triangle O A B$ is $7+2 \sqrt{10}$.