13. (16 points) As shown in Figure 2, given

semicircle $C: x^{2}+y^{2}=1(y \geqslant 0)$ and point $A(2,0)$, $B$ is any point on the semicircle, and an equilateral triangle $\triangle A B M$ is constructed with $A B$ as one side. When is the area of quadrilateral $O A M B$ maximized? And find this maximum value.
Three, 13. Let $\angle A O B=\theta(0<\theta<\pi)$.
Since $O B=1, O A=2$, we have
$$
A B^{2}=1^{2}+2^{2}-2 \times 1 \times 2 \cos \theta=5-4 \cos \theta \text {. }
$$

Therefore, $S_{\text {quadrilateral } O A M B}$
$$
\begin{array}{l}
=\frac{1}{2} \times 1 \times 2 \sin \theta+\frac{\sqrt{3}}{4}(5-4 \cos \theta) \\
=\sin \theta-\sqrt{3} \cos \theta+\frac{5 \sqrt{3}}{4} \\
=2 \sin \left(\theta-\frac{\pi}{3}\right)+\frac{5 \sqrt{3}}{4} .
\end{array}
$$

Since $0<\theta<\pi$, when $\theta=\frac{5 \pi}{6}$, $S_{\text {quadrilateral } O A M B}$ achieves its maximum value $2+\frac{5 \sqrt{3}}{4}$.

Thus, when $\angle A O B=\frac{5 \pi}{6}$, the area of quadrilateral $O A M B$ reaches its maximum value $2+\frac{5 \sqrt{3}}{4}$.