Question

Find the volume of a right circular cone that has a height of $19.2\mathrm{ft}$ and a base with a circumference of $7.6\mathrm{ft}$. Round your answer to the nearest tenth of a cubic foot.

Answer 1

Step-1: Find radius of the base of cone:

Given: Height of cone $=19.2\mathrm{ft}$

Circumference of base $=7.6\mathrm{ft}$

Let the radius of cone be $\mathrm{r}\mathrm{ft}$

We know that circumference of circle $=2\mathrm{\pi r}$

$\Rightarrow$$2\mathrm{\pi r}=7.6$

$\Rightarrow$ $\mathrm{r}=\frac{7.6}{2\mathrm{\pi }}\mathrm{ft}$

Step-2: Find the volume:

Volume of the cone $=\frac{1}{3}\mathrm{\pi }{\left(\mathrm{r}\right)}^2\mathrm{h}$

So, volume of the given cone$=\frac{1}{3}\mathrm{\pi }{\left(\frac{7.6}{2\mathrm{\pi }}\right)}^2\left(19.2\right)$

$=\frac{1}{3}\times \frac{{\left(7.6\right)}^2}{4\mathrm{\pi }}\times \left(19.2\right)$

$=\frac{1}{3}\times \frac{57.76}{4\left(3.14\right)}\times \left(19.2\right)$ $\left[\because \mathrm{\pi }=3.14\right]$

$=29.431{\text{ft}}^3$

$\approx 29.4{\mathrm{ft}}^3$ [Rounded off to the nearest tenth]

Hence, volume of the given right circular cone is $29.4{\mathrm{ft}}^3$, rounded off to the nearest tenth.