If the area of a regular hexagon $A B C D E F$ is $54 \sqrt{3} \mathrm{~cm}^{2}$ and $A B=x \mathrm{~cm}, A C=y \sqrt{3} \mathrm{~cm}$,
G9.1 find $x$.
G9.2 find $y$.
The hexagon can be cut into 6 identical equilateral triangles
$$
\begin{array}{l}
6 \cdot \frac{1}{2} \cdot\left(x^{2} \cdot \sin 60^{\circ}\right)=54 \sqrt{3} \\
\Rightarrow x=6 \\
\angle A B C=120^{\circ} \\
\begin{aligned}
A C^{2} & =\left(x^{2}+x^{2}-2 x^{2} \cos 120^{\circ}\right) \mathrm{cm}^{2} \\
\quad & =\left[6^{2}+6^{2}-2(6)^{2} \cdot\left(-\frac{1}{2}\right)\right] \mathrm{cm}^{2} \\
\quad & =3 \times 6^{2} \mathrm{~cm}^{2}
\end{aligned} \\
y \sqrt{3}=6 \sqrt{3} \\
\Rightarrow y=6
\end{array}
$$