10. (12 points) In the figure, segments $A E$ and $F G$ divide rectangle $A B C D$ into four parts; it is known that the areas of two of these parts are 2 square centimeters and 11 square centimeters, and $E$ is the midpoint of $B C$, $O$ is the midpoint of $A E$; then the area of rectangle $A B C D$ is $\qquad$ square centimeters.
【Solution】Solution: According to the analysis, extend $A E, D C$ to intersect at point $H$, then $A F O G H$ is a hourglass model, and $A B E C H$ is also an hourglass model,

Since $E$ is the midpoint of $B C$, we have $A E: E H=B E: E D=1: 1$, and since $O$ is the midpoint of $A E$, then $A O: O H=1$ : 3,

Therefore, in the hourglass model $A F O G H$, we have $S_{\triangle \mathrm{AOF}}: S_{\triangle G O H}=1^{2}: 3^{2}=1: 9$, so, $S_{\triangle G O H}=2 \times 9=18$ (square centimeters), then $S_{\triangle C E H}=18-11=7$ (square centimeters),

The area of the rectangle is exactly 4 times the area of $\triangle E C H$, so $S_{\triangle A B C D}=4 S_{\triangle A B E}=4 S_{\triangle C E H}=4 \times 7=28$ (square centimeters) Therefore, the answer is: 28.