9. As shown in the figure, $\mathrm{ABCD}$ and $\mathrm{EGFO}$ are both squares, where point $\mathrm{O}$ is the center of square $\mathrm{ABCD}$, and $\mathrm{EF} / / \mathrm{BC}$. If the lengths of BC and EF are both positive integers, and the area of quadrilateral BCEF is 3.25, then $S_{A B C D}-S_{E G F O}=$

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【Analysis】As shown in the figure, draw $\mathrm{ON}$ perpendicular to $\mathrm{EF}$ at $\mathrm{N}$, intersecting $\mathrm{BC}$ at $\mathrm{M}$. It is easy to see that $\mathrm{ON}$ is the altitude from the hypotenuse $\mathrm{EF}$ of the isosceles right triangle $\mathrm{OEF}$, and $\mathrm{OM}$ is the altitude from the hypotenuse $\mathrm{BC}$ of the isosceles right triangle $\mathrm{OBC}$. Let $O N=a$ and $O M=b$, then we have $E F=2 O M=2 a, B C=2 O M=2 b$. $S_{\text {BCEF }}=(B C+E F) \times M N \div 2=(2 b+2 a) \times(a-b) \div 2=(a+b) \times(a-b)=a^{2}-b^{2}$. Therefore, $a^{2}-b^{2}=3.25 \Rightarrow E F^{2}-B C^{2}=(2 a)^{2}-(2 b)^{2}=4\left(a^{2}-b^{2}\right)=13$. Since the lengths of $\mathrm{BC}$ and $\mathrm{EF}$ are both integers, we have $E F=7, B C=6$. Therefore, $S_{A B C D}-S_{E G F O}=B C^{2}-E F^{2} \div 2=11.5$.