11.002. Calculate the volume of a regular tetrahedron if the radius of the circle circumscribed around its face is $R$.
Solution.

For a tetrahedron with equal edges, denote the edges by \(a\). The radius of the circle circumscribed around the base of the tetrahedron (which is an equilateral triangle) is \(R = \frac{a \sqrt{3}}{3}\), from which \(a = R \sqrt{3}\). The height of the tetrahedron projects onto the center of the circle circumscribed around the base (Fig. 11.4). Therefore, \(S H = \sqrt{S A^{2} - A H^{2}} = \sqrt{a^{2} - R^{2}} = \sqrt{3 R^{2} - R^{2}} = R \sqrt{2}\). The area of the base is \(S_{\text{base}} = \frac{a^{2} \sqrt{3}}{4} = \frac{3 \sqrt{3} R^{2}}{4}\). Thus, the volume is

\[
V = \frac{1}{3} S_{\text{base}} \cdot S H = \frac{3 \sqrt{3}}{4} R^{2} \cdot R \sqrt{2} \cdot \frac{1}{3} = \frac{R^{3} \sqrt{6}}{4}
\]

Answer: \(\frac{R^{3} \sqrt{6}}{4}\).