6. A right triangle $\triangle A B C$ with legs measuring 18 cm and 10 cm, and a right triangle $\triangle A D E$ with legs measuring 14 cm and 4 cm are placed as shown. $M$ is the midpoint of $A E$. The area of $\triangle A C M$ is $\qquad$ square centimeters.
【Solution】Solution: $S_{\triangle A B C}+S_{\triangle A D E}=(18 \times 10+14 \times 4) \div 2=(180+56) \div 2=118$ (square centimeters) According to $\triangle D E F \sim \triangle B C F$ we get:
$$
\begin{array}{l}
B F: D F=B C: D E=10: 4=2.5 \\
B F+D F=4,
\end{array}
$$

Solving, we get $D F=\frac{8}{7}, B F=\frac{20}{7}$,
$$
\begin{array}{l}
\because A M=M E, \\
\therefore S_{\text {ming }}=\left(S_{\triangle A B C}+S_{\triangle A D E}+S_{\triangle D E F}-S_{\triangle B C F}\right) \div 2=\left(118+0.5 \times \frac{8}{7} \times 4-0.5 \times \frac{20}{7} \times 10\right) \div 2=
\end{array}
$$
53 (square centimeters)
Therefore, the answer is 53.