G10.2 The medians $A L, B M, C N$ of $\triangle A B C$ meet at $G$. If the area of $\triangle A B C$ is $54 \mathrm{~cm}^{2}$ and the area of $\triangle A N G$ is $x \mathrm{~cm}^{2}$. Find $x$.
$\begin{array}{l}\because B L: L C=1: 1 \\ \therefore \text { area of } \triangle A B L=\frac{1}{2} \cdot 54 \mathrm{~cm}^{2}=27 \mathrm{~cm}^{2} \\ \because A G: G L=2: 1 \\ \therefore \text { area of } \triangle A B G=\frac{2}{3} \cdot 27 \mathrm{~cm}^{2}=18 \mathrm{~cm}^{2} \\ \because A N: N B=1: 1 \\ \therefore \text { area of } \triangle A N G=\frac{1}{2} \cdot 18 \mathrm{~cm}^{2}=9 \mathrm{~cm}^{2} \\ x=9\end{array}$