Let $ABC$ be a given equilateral triangle with side $a$, $O$ be its center, and the axis of rotation pass through point $O$ parallel to $BC$. Then the axial section of the resulting solid of revolution is the union of triangle $ABC$ and an equal triangle $A1B1C1$ with the same center $O$ and side $B1C1$ parallel to $BC$. The area $S$ of this section is equal to the sum of the areas of triangle $ABC$ and three equilateral triangles with sides $\frac{a}{3}$.

Therefore,

$$
S=\frac{a^{2} \sqrt{3}}{4}+3 \cdot \frac{1}{9} \cdot \frac{a^{2} \sqrt{3}}{4}=\frac{a^{2} \sqrt{3}}{4}+\frac{1}{3} \cdot \frac{a^{2} \sqrt{3}}{4}=\frac{a^{2} \sqrt{3}}{3}
$$

## Answer

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

Note: There is a minor correction in the translation. The formula for the area of an equilateral triangle is $\frac{a^2 \sqrt{3}}{4}$, not $\frac{a^2 \sqrt{3}}{1}$. The final result remains the same.
Find the area of the axial section of the body obtained by rotating an equilateral triangle with side $a$ around a line passing through its center and parallel to one of its sides.

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