6. (10 points) Square $ABCD$ and rectangle $BEFG$ are placed as shown in the figure, with $AG=CE=2$ cm. The area of square $ABCD$ is larger than the area of rectangle $BEFG$ by $\qquad$ square centimeters.
【Solution】Solution: According to the analysis, the common part in the figure is rectangle $G H C B$, so:
the area of square $A B C D$ - the area of rectangle $B E F G$ = the area of rectangle $A D H G$ - the area of rectangle $E F H C$ = $A G \times A D - C E \times C H = 2 \times A D - 2 \times C H = 2 \times (A D - C H) = 2 \times (C D - C H) = 2 \times D H = 2 \times 2 = 4$ (square centimeters).

Therefore, the answer is: 4.