9. (10 points) In the figure, $AB$ is the diameter of circle $O$, 6 cm long, and square $BCDE$ has one vertex $E$ on the circumference of the circle, $\angle ABE=45^{\circ}$. The difference between the area of the non-shaded part of circle $O$ and the area of the non-shaded part of square $BCDE$ is $\qquad$ square centimeters (take $\pi=3.14$)
【Analysis】Connect $E O$, the area of the non-shaded part of circle $O$ - the area of the non-shaded part of square $B C D E$ = (the area of the non-shaded part of circle $O$ + the area of the shaded part) - (the area of the non-shaded part of square $B C D E$ + the area of the shaded part) leads to $B E^{2}=r^{2}=(6 \div 2)^{2} \times 2$, then substitute the data into the previous relationship formula to solve.

【Solution】Solution:
As shown in the figure, connect $E O, S_{\text {square }}=E B \times E B=E O^{2}+B O^{2}=(6 \div 2)^{2} \times 2=18 \mathrm{~cm}^{2}$
Therefore, the difference between the area of the non-shaded part of circle $O$ and the area of the non-shaded part of square $B C D E$: $\pi \times(6 \div 2)^{2}-18=10.26$ (square centimeters);

Answer: The difference between the area of the non-shaded part of circle $O$ and the area of the non-shaded part of square $B C D E$ is 10.26 square centimeters.

The answer is: 10.26.