Mensuration Practice Test 4

Edge of a cube inscribed in a sphere
= $\frac{\mathrm{Diameter}}{\sqrt{3}}$ = $\frac{12}{\sqrt{3}}$
= $4\sqrt{3}$ cm
Volume of the cube = ${\left(\mathrm{side}\right)}^3$
= ${\left(4\sqrt{3}\right)}^3$
= $576\sqrt{3}\mathit{c}{\mathit{m}}^2$
Volume of the sphere = $\frac{4}{3}\mathit{\pi }{\mathit{r}}^2$
= $\frac{4}{3}\times 3.14\times 3^2$ = $113.04\mathit{c}{\mathit{m}}^3$
Edge of a cube = $\frac{\mathrm{Diameter}\mathrm{of}\mathrm{the}\mathrm{sphere}}{\sqrt{3}}$
= $\frac{6}{\sqrt{3}}$ = $2\sqrt{3}$ cm
Volume of the cube = ${\left(\mathrm{side}\right)}^3$
= ${\left(2\sqrt{3}\right)}^3$
= $24\sqrt{3}$
= 24 × 1.73
= 41.57 $\mathit{c}{\mathit{m}}^3$
Required volume = 113.04 - 41.57
= 71.47 $\mathit{c}{\mathit{m}}^3$