11.171. A cube is inscribed in a hemisphere of radius $R$ such that four of its vertices lie on the base of the hemisphere, while the other four vertices are located on its spherical surface. Calculate the volume of the cube.
Solution.

Consider the section of the specified hemisphere and cube by a plane passing through the opposite lateral edges $A A_{1}$ and $C C_{1}$ of the cube (Fig. 11.65), $O$ - the center of the hemisphere and lies on the diagonal $A C$ of the base of the cube. Let the edge of the cube $A A_{1}=x$. Then $O A=\frac{x \sqrt{2}}{2}$ and from $\triangle A_{1} A O\left(\angle A_{1} A O=90^{\circ}\right): O A_{1}=\sqrt{A A_{1}^{2}+O A^{2}}=\sqrt{x^{2}+\frac{x^{2}}{2}}=x \sqrt{\frac{3}{2}}$. Since $O A_{1}=R$, then $x \sqrt{\frac{3}{2}}=R ; x=R \sqrt{\frac{2}{3}}$ and the volume of the cube $V=x^{3}=\frac{2}{3} R^{3} \sqrt{\frac{2}{3}}=\frac{2 R^{3} \sqrt{6}}{9}$. Answer: $\quad \frac{2 R^{3} \sqrt{6}}{9}$.