Let's inscribe a regular hexagon and rotate it around its axes of symmetry. How do the surface areas of the resulting solids of revolution compare?
Solution. A regular hexagon has two types of axes of symmetry.

First, consider the one that passes through the midpoints of two opposite sides. When the hexagon is rotated around this axis, two congruent frustums of cones are formed. The surface area in question consists of the lateral surface areas of two frustums and the areas of two circles (the base circle area is not considered).

Let the side of the hexagon be of unit length, then the radius of the top circle $r=\frac{1}{2}$, the radius of the base circle $R=1$, and the slant height of the frustum $a=1$. The area of the top circle is $\frac{1}{4} \pi$, and the lateral surface area of the frustum is $\frac{3}{2} \pi$.

![](https://cdn.mathpix.com/cropped/2024_05_02_a07e65417d65f018022ag-1.jpg?height=325&width=336&top_left_y=453&top_left_x=884)

The sought surface area:

$$
F_{1}=2 \cdot \frac{1}{4} \pi+2 \cdot \frac{3}{2} \pi=\frac{7}{2} \pi
$$

In the other case, the axis of symmetry passes through two opposite vertices of the hexagon. In this case, two congruent cones and a cylinder are formed when rotated.

Consider the isosceles triangle $ABC$. Let the midpoint of $AC$ be $D$. The triangle $BDC$ is half of a regular triangle with a side length of 1 unit, so $r=\frac{\sqrt{3}}{2}$.

The surface area is the sum of the lateral surface areas of the two cones and the lateral surface area of the cylinder:

$$
\begin{aligned}
T_{\text {cones }} & =2 r \pi a=2 \cdot \frac{\sqrt{3}}{2} \pi \cdot 1=\sqrt{3} \pi \\
T_{\text {cylinder }} & =2 r \pi m=\sqrt{3} \pi
\end{aligned}
$$

The sum of the two surface areas is $F_{2}=2 \sqrt{3} \pi$.

![](https://cdn.mathpix.com/cropped/2024_05_02_a07e65417d65f018022ag-1.jpg?height=583&width=509&top_left_y=1470&top_left_x=797)

The ratio of the surface areas:

$$
\frac{F_{1}}{F_{2}}=\frac{\frac{7}{2} \pi}{2 \sqrt{3} \pi}=\frac{7}{4 \sqrt{3}} \approx 1.0104
$$

The surface areas of the two solids of revolution are approximately equal.