甽 2. As shown in the figure, in $\triangle ABC$, it is known that $M, N$ are on sides $AC, BC$ respectively, and $BM$ intersects $AN$ at $O$. If $S_{\triangle OMA}=3 \text{~cm}^{2}, S_{\triangle OAB}$ $=2 \text{~cm}^{2}, S_{\triangle OBN}=1 \text{~cm}^{2}$.

Find $S_{\triangle CMN}$. (1990, Shanghai Junior High School Mathematics Competition)
Analysis From the figure, we know that $\triangle A B C$ is composed of $\triangle C M A, \triangle O A B$, $\triangle O B N, \triangle O M N, \triangle C M N$, and the areas of the latter triangles.

First, $\because \triangle A R O$ and $\triangle A O M$ have the same height, $\therefore S_{\triangle O M A}$ 1 $S_{\triangle, O B B}=3: 2=M O: O B$. Also, since $\triangle N B O$ and $\triangle N M O$ have the same height, $\therefore S_{\triangle O B N}: S_{\triangle O M N}=O B: M O=2: 3 \Rightarrow S_{\triangle O N N}=$ $\frac{3}{2}\left(\mathrm{~cm}^{2}\right)$. Thus, $S_{\triangle A M N}=4 \frac{1}{2}\left(\mathrm{~cm}^{2}\right)$. Then, from $S_{\triangle C M N}$, $S_{\triangle A M N}=\frac{C M}{A M}$ and $\frac{C M}{A M}=S_{\triangle B C M}: S_{\triangle A B M}=\left(1+\frac{3}{2}+\right.$ $\left.S_{\triangle C M N}\right):(2+3)$, we get $S_{\triangle C M N}=22.5\left(\mathrm{~cm}^{2}\right)$.