10.112. Calculate the ratio of the areas of a square, an equilateral triangle, and a regular hexagon inscribed in the same circle.
Solution.

Let $R$ be the radius of the circle. Then the side of the inscribed regular triangle is $a_{3}=R \sqrt{3}$ and $S_{3}=\frac{a_{3}^{2} \sqrt{3}}{4}=\frac{3 R^{2} \sqrt{3}}{4}$. Next, the side of the square is $a_{4}=R \sqrt{2}$ and $S_{4}=a_{4}^{2}=2 R^{2}$, and finally, the side of the inscribed regular hexagon is $a_{6}=R$ and $S_{6}=\frac{6 R^{2} \sqrt{3}}{4}=\frac{3 R^{2} \sqrt{3}}{2}$. Therefore, $S_{4}: S_{3}: S_{6}=8: 3 \sqrt{3}: 6 \sqrt{3}$.

Answer: $\quad 8: 3 \sqrt{3}: 6 \sqrt{3}$.