II. (This question is worth 20 points) Given the line $l: y=\sqrt{3} x+4$, the moving circle $O: x^{2}+y^{2}=r^{2}(1<r<2)$, and the rhombus $A B C D$ with one interior angle of $60^{\circ}$, where vertices $A, B$ lie on the line $l$, and vertices $C, D$ lie on the circle $O$. Find the range of the area $S$ of the rhombus $A B C D$ as $r$ varies.
Solution: Since quadrilateral $ABCD$ has an internal angle of $60^{\circ}$, $\triangle ACD$ or $\triangle BCD$ is an equilateral triangle. Without loss of generality, let $\triangle ACD$ be the equilateral triangle, as shown in Figure 2.
Since the distance from the center $O$ to the line $l$ is $2>r$, the line $l$ is separated from the circle $O$. …․5 points. Let the equation of line $CD$ be $y=\sqrt{3} x+b$, then the distance between line $l$ and $CD$ is $d=\frac{|b-4|}{2}$.
Also, the distance from the center $O$ to the line $CD$ is $\frac{|b|}{2}$, so $|CD|=2 \sqrt{r^{2}-\left(\frac{|b|}{2}\right)^{2}}$ $=\sqrt{4 r^{2}-b^{2}}$.
From $d=\frac{\sqrt{3}}{2}|CD|$, we get $\frac{|b-4|}{2}=\frac{\sqrt{3}}{2} \sqrt{4 r^{2}-b^{2}}$.
Simplifying, we get $b^{2}-2 b+4=3 r^{2}$.
Since $1<r<2$, we have $3<b^{2}-2 b+4<12$.
Solving, we get $-2<b<1$, or $1<r<4$.
Also, $S=2 S_{\triangle \mid CD}=2 \times \frac{\sqrt{3}}{4}|CD|^{2}=2 \times \frac{\sqrt{3}}{4} \times\left(\frac{2}{\sqrt{3}}\right)^{2} d^{2}=\frac{\sqrt{3}}{6}(b-4)^{2}$.
15 points
Since the function $S=\frac{\sqrt{3}}{6}(b-4)^{2}$ is monotonically increasing on $(-2,1)$ and $(1,4)$, the range of the area $S$ of quadrilateral $ABCD$ is $\left(0, \frac{3}{2} \sqrt{3}\right) \cup\left(\frac{3}{2} \sqrt{3}, 6 \sqrt{3}\right)$.
20 points