Solution.

The desired area $S=12 S_{\triangle O A B}$, where $O A, O B$ are drawn to the adjacent vertices of a regular dodecagon (Fig. 10.97). The side of the square is $a$, so $O A=O B=\frac{a}{\sqrt{2}}$.

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Fig. 10.97

Draw the height $B C \perp O A$. Then $S_{\triangle O A B}=\frac{1}{2} O A \cdot B C$, where $\angle A O B=\frac{2 \pi}{12}=\frac{\pi}{6}$.

We get $S_{\triangle O A B}=\frac{1}{2} \frac{a}{\sqrt{2}} \cdot \frac{1}{2} \frac{a}{\sqrt{2}}=\frac{a^{2}}{8} \Rightarrow S=\frac{12}{8} a^{2}=\frac{3}{2} a^{2}$. Answer: $\frac{3 a^{2}}{2}$.

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Fig. 10.98

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Fig. 10.99