10.043. A regular triangle is inscribed in a circle with a diameter equal to $\sqrt{12}$. On its height as a side, another regular triangle is constructed, in which a new circle is inscribed. Find the radius of this circle.

![](https://cdn.mathpix.com/cropped/2024_05_22_fa9db84b44b98ec0a5c7g-576.jpg?height=394&width=372&top_left_y=79&top_left_x=252)

Fig. 10.43

![](https://cdn.mathpix.com/cropped/2024_05_22_fa9db84b44b98ec0a5c7g-576.jpg?height=336&width=326&top_left_y=140&top_left_x=777)

Fig. 10.44
## Solution.

Given $R=0.5 \sqrt{12}$. The side $AB$ of the inscribed regular triangle (Fig. 10.43) is $R \sqrt{3}$, i.e., $0.5 \sqrt{12} \cdot \sqrt{3}=3$. We will find the side $CD$ of the new triangle: $a=\sqrt{3^{2}-1.5^{2}}=1.5 \sqrt{3}$. Since the radius $r$ of the circle inscribed in it is $a \sqrt{3} / 6$, we finally get $r=1.5 \sqrt{3} \cdot \sqrt{3} / 6=3 / 4$.

Answer: $3 / 4$.