Analysis:
First, analyze the regular hexagon. As shown in the left figure, the area of a regular hexagon can be expressed as $6 S$. It is easy to see that the area of triangle $D E F$ is $S$, and the area of triangle $C D F$ is $2 S$.

As shown in the right figure, connect $D F$ and $C F$. Let the area of the regular hexagon $A B C D E F$ be $6 S$, then the area of triangle $D E F$ is $S$, and the area of triangle $C D F$ is $2 S$.

Since $S_{\triangle F D M}: S_{\triangle F C M}=D M: C M=2: 3$
Therefore, $S_{\triangle F D M}=\frac{2}{5} S_{\triangle F C D}=\frac{2}{5} \times 2 S=\frac{4}{5} S$
Thus, $S_{\text {quad } F E D M}=S_{\triangle F E D}+S_{\triangle F D M}=S_{\text {quad } F E D N}+S_{\triangle N D M}$
That is, $S+\frac{4}{5} S=4+2$, solving for $S$ gives $S=\frac{10}{3}$
Therefore, $S_{A B C D E F}=6 S=20\left(\mathrm{~cm}^{2}\right)$
Therefore, $S_{\text {shaded }}=20-2-3-4=11\left(\mathrm{~cm}^{2}\right)$
【Question 13】As shown in the figure, three line segments divide the regular hexagon into four parts. It is known that the areas of three parts are $2$, $3$, and $4$ square centimeters, respectively. What is the area of the fourth part (the shaded area in the figure)? $\qquad$ square centimeters.