9. In the right figure, $AB$ is the diameter of circle $O$, 6 cm long, and one vertex $E$ of square $BCDE$ is on the circumference of the circle, $\angle \mathrm{ABE}=45^{\circ}$. 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: Graphic area, difference remains constant problem.
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) $=\mathrm{S}$ four $\mathrm{S}$

The key is to find the area of the square. As shown in the figure, connect $\mathrm{E} O, \mathrm{~S}_{\mathrm{E}}=\mathrm{EB} \times \mathrm{EB}=\mathrm{E} 0^{2}+B 0^{2}=(6 \div 2)^{2} \times 2=18 \mathrm{~cm}^{2}$, so the difference between the area of the non-shaded part of circle $O$ and the area of the non-shaded part of square $B C D E$:
$$
\pi \times(6 \div 2)^{2}-18=10.28 \mathrm{~cm}^{2} .
$$