12. (10 points) Two squares $A B C D$ and $B E F G$ with side lengths of $8 \mathrm{~cm}$ and $6 \mathrm{~cm}$ respectively are placed side by side as shown in the figure. Connecting $D E$ intersects $B G$ at $P$. What is the area of the shaded region $A P E G$ in the figure?
【Analysis】First, we need to find the relationship between the shaded part and the side length of the known square. We can transform $\triangle A P G$ into $\triangle D P G$ with the same base and height, and then connect it to the square with a side length of 6 using the principle of equal area transformation.

【Solution】Solution: According to the problem, we know: Move $\triangle A P G$ to $\triangle D P G$ (as shown in the middle figure above), connect $D B, D B$ is parallel to $G E$. The area of $\triangle D G E$ is equal to the area of $\triangle B G E$ (as shown in the right figure above).
$$
S_{\text {light }}=6 \times 6 \div 2=18 \mathrm{~cm}^{2} \text {. }
$$

Answer: The area of the shaded part $A P E G$ is $18 \mathrm{~cm}^{2}$.