Volumes Practice Test 3

Suppose the base radius and the height of the right circular cone are 3x m and 5x m respectively.
∴ Volume of the cone = $\frac{1}{3}\mathit{\pi }{\mathit{r}}^2\mathit{h}$ = $\frac{1}{3}\mathit{\pi }{\left(3\mathit{x}\right)}^2\times 5\mathit{x}$ cu m
Now, $\frac{1}{3}\mathit{\pi }\times 9{\mathit{x}}^2\times 5\mathit{x}$ = 120π
⇒ ${\mathit{x}}^3$ = $\frac{120\times 3}{9\times 5}$
⇒ ${\mathit{x}}^3$ = 8 ⇒ ${\mathit{x}}^3$ = ${\left(2\right)}^3$ ⇒ x = 2 m
∴ The radius and the height of the cone will be 3 × 2 = 6 m and 5 × 2 = 10 m respectively.
∴ Slant height of the cone = $\sqrt{{\mathit{r}}^2+{\mathit{h}}^2}$ = $\sqrt{{\left(6\right)}^2+{\left(10\right)}^2}$ = $\sqrt{36+100}$ = $\sqrt{136}$ = $2\sqrt{34}$ m.