10.308. Given a square with side $a$. On each side of the square, outside of it, a trapezoid is constructed such that the upper bases of these trapezoids and their lateral sides form a regular dodecagon. Calculate its area.
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 $O A=O B=\frac{a}{\sqrt{2}}$.

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0784.jpg?height=496&width=523&top_left_y=935&top_left_x=740)

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: $\frac{3 a^{2}}{2}$.

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0785.jpg?height=414&width=648&top_left_y=104&top_left_x=96)

Fig. 10.98

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0785.jpg?height=440&width=448&top_left_y=79&top_left_x=753)

Fig. 10.99