Mensuration Practice Test 4

Volume of sphere = $\frac{4}{3}\times \frac{22}{7}\times 12\times 12\times 12$
= $\frac{50688}{7}$ = 7241.14 $\mathit{c}{\mathit{m}}^3$
Volume of two smaller spheres
= $\frac{4}{3}\times \frac{22}{7}\times 6\times 6\times 6$ + $\frac{4}{3}\times \frac{22}{7}\times 8\times 8\times 8$
= 905.14 + 2,145.52
= 3,050.66 $\mathit{c}{\mathit{m}}^3$
Volume of the third sphere
= 7,241.14 − 3,050.66 = 4,190.48 $\mathit{c}{\mathit{m}}^3$
Radius of the sphere = $\frac{1}{3}\mathit{\pi }{\mathit{r}}^3$ = 4190.48
${\mathit{r}}^3$ = $4190.48\times \frac{3}{4}\times \frac{7}{22}$
${\mathit{r}}^3$ = 4190.48 × $\frac{3}{4}\times \frac{7}{22}$
${\mathit{r}}^3$ = 1000
r = 10 cm