10.142. A circle is inscribed in an equilateral triangle, and a regular hexagon is inscribed in the circle. Find the ratio of the areas of the triangle and the hexagon.
## Solution.

Let the side of the equilateral triangle be $a$. Then its area $S_{1}=\frac{a^{2} \sqrt{3}}{4} \cdot$ The radius of the circle inscribed in the triangle, $r=\frac{a \sqrt{3}}{6}$.

![](https://cdn.mathpix.com/cropped/2024_05_22_fa9db84b44b98ec0a5c7g-623.jpg?height=352&width=534&top_left_y=96&top_left_x=105)

Fig. 10.128

![](https://cdn.mathpix.com/cropped/2024_05_22_fa9db84b44b98ec0a5c7g-623.jpg?height=318&width=514&top_left_y=133&top_left_x=731)

Fig. 10.129

It will be equal to the side of the hexagon inscribed in this circle: $a_{6}=\frac{a \sqrt{3}}{6} \cdot$ And the radius of the circle inscribed in this hexagon:

$$
r=\frac{a_{6}}{2 \operatorname{tg} \frac{180^{\circ}}{6}}=\frac{a_{6} \sqrt{3}}{2}=\frac{a}{4}
$$

The area of the hexagon

$$
S_{2}=\frac{6 a_{5} r}{2}=\frac{a^{2} \sqrt{3}}{8} ; \frac{S_{1}}{S_{2}}=\frac{a^{2} \sqrt{3} \cdot 8}{4 a^{2} \sqrt{3}}=2
$$

Answer: 2.