We inscribed the regular hexagon $A B C D E F$ in a circle, then drew semicircles outward over the chords $A B, B D, D E$, and $E A$. Calculate the ratio of the total area of the 4 lens-shaped (bounded by two circular arcs) figures that formed to the area of the hexagon.

Let's start by calculating the area of the regular hexagon. The area of a regular hexagon with side length $s$ is given by:
\[ T_{\text{hexagon}} = \frac{3 \sqrt{3}}{2} s^2 \]

Next, we need to calculate the area of the lens-shaped figures. Each lens is formed by the intersection of two semicircles. The radius of each semicircle is equal to the side length of the hexagon, $s$. The area of each semicircle is:
\[ T_{\text{semicircle}} = \frac{1}{2} \pi s^2 \]

The area of the lens-shaped figure can be found by subtracting the area of the equilateral triangle formed by the two radii and the chord from the area of the semicircle. The area of an equilateral triangle with side length $s$ is:
\[ T_{\text{triangle}} = \frac{\sqrt{3}}{4} s^2 \]

Thus, the area of one lens-shaped figure is:
\[ T_{\text{lens}} = T_{\text{semicircle}} - T_{\text{triangle}} = \frac{1}{2} \pi s^2 - \frac{\sqrt{3}}{4} s^2 = \left( \frac{\pi}{2} - \frac{\sqrt{3}}{4} \right) s^2 \]

Since there are 4 such lens-shaped figures, the total area of the lens-shaped figures is:
\[ T_{\text{total lenses}} = 4 \left( \frac{\pi}{2} - \frac{\sqrt{3}}{4} \right) s^2 = \left( 2\pi - \sqrt{3} \right) s^2 \]

Finally, the ratio of the total area of the lens-shaped figures to the area of the hexagon is:
\[ \text{Ratio} = \frac{T_{\text{total lenses}}}{T_{\text{hexagon}}} = \frac{\left( 2\pi - \sqrt{3} \right) s^2}{\frac{3 \sqrt{3}}{2} s^2} = \frac{2\pi - \sqrt{3}}{\frac{3 \sqrt{3}}{2}} = \frac{2(2\pi - \sqrt{3})}{3 \sqrt{3}} = \frac{4\pi - 2\sqrt{3}}{3 \sqrt{3}} = \frac{4\pi - 2\sqrt{3}}{3 \sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}} = \frac{4\pi \sqrt{3} - 6}{9} = \frac{4\pi \sqrt{3} - 6}{9} \]

Thus, the ratio is:
\[ \boxed{\frac{4\pi \sqrt{3} - 6}{9}} \]
A 4-leaf clover-shaped figure's total area is obtained by subtracting the area of the circle with diameter $AD$ (the circle circumscribed around the hexagon) from the sum of the area of the rectangle $ABDE$ and the areas of the semicircles.

![](https://cdn.mathpix.com/cropped/2024_05_02_ddd4af87fe91348bdc87g-1.jpg?height=412&width=423&top_left_y=274&top_left_x=840)

The sum of the areas of the mentioned semicircles is equal to the sum of the areas of the circles with diameters $AB$ and $BD$, which is $\left(AB^{2} + BD^{2}\right) \cdot \pi / 4$. According to the Pythagorean theorem, this is equal to the area of the circle with diameter $AD$, which is $AD^{2} \pi / 4$. Therefore, the total area of the 4-leaf clover-shaped figure is equal to the area of the rectangle $ABDE$.

The area of the rectangle $ABDE$ is four times the area of the triangle $ABO$ (where $O$ is the center of the hexagon). The area of the hexagon is six times the area of the triangle $ABO$. Therefore, the ratio of the area of the 4-leaf clover-shaped figure to the area of the hexagon is $4: 6 = 2: 3$.