11. (12 points) In the figure, the area of rectangle $A B C D$ is 2013 square centimeters. $\triangle A O D$, $\triangle B O C$, $\triangle A B E$, $\triangle B C F$, $\triangle C D G$, $\triangle A D H$ are all equilateral triangles, and $M$, $N$, $P$, $Q$ are the centers of $\triangle A B E$, $\triangle B C F$, $\triangle C D G$, $\triangle A D H$ respectively. The area of the shaded part is $\qquad$ square centimeters.
【Solution】Solution: As shown in the figure, connect $A M$, $O M$, $O Q$, and let $O Q$ intersect $A D$ at point $K$,
$\because Q$ is the center of the equilateral $\triangle A D H$, and $\triangle A O D$ is identical to $\triangle A D H$,
$\therefore Q K=\frac{1}{3} O K \therefore O Q=\frac{4}{3} O K, \triangle O Q M$ has a height of $O Q$ and a base of $O M, \triangle A O D$ has a height of $O K$,
and a base of $A D$, since $O M=A D, \therefore S_{\triangle O Q M}=\frac{4}{3} S_{\triangle M O D}$;
also, $S_{\triangle O Q M}=\frac{1}{4} S_{\text {shaded}}, S_{\triangle A O D}=\frac{1}{4} S_{A B C D}, \therefore S_{\text {shaded}}=\frac{4}{3} S_{\text {ABCD}}=\frac{4}{3} \times 2013=2684$ (square centimeters).

Therefore, the answer is: 2684.