Question

The volume of a right circular cone is $9856{\text{cm}}^3$. If the diameter of the base is $28\text{cm}$, find the

(1) height of the cone

(2) slant height of the cone

(3) curved surface area of the cone

Answer 1

Step-1: Height of the cone:

We know that volume of the cone $\mathbf{=}\frac{\mathbf{1}}{\mathbf{3}}\mathbf{\pi }{\mathbf{\text{r}}}^{\mathbf{2}}\mathbf{\text{h}}$

Here volume $=9856{\text{cm}}^3$

Diameter of the base $=28\text{cm}$

Radius of the base $=\frac{\text{diameter}}{2}=\frac{28}{2}=14\text{cm}$

So $\frac{1}{3}{\mathrm{\pi r}}^2\mathrm{h}=9856$

After putting the values we get

$$\begin{gathered} \frac{1}{3}\times \frac{22}{7}\times 14\times 14\times \text{h}=9856\left[\because \pi =\frac{22}{7}\right] \\ \text{h=}\frac{9856\times 3\times 7}{22\times 14\times 14}\text{=48}cm \end{gathered}$$

So height of the cube is $=48\text{cm}$

Step-2: Slant height of the cone:

Let $\text{l}$ is the slant height of the cone and we know that

$$\begin{gathered} {\mathbf{\text{l}}}^{\mathbf{2}}{\mathbf{\text{=h}}}^{\mathbf{2}}{\mathbf{\text{+r}}}^{\mathbf{2}} \\ \text{l=}\sqrt{{\text{h}}^2{\text{+r}}^2}\text{=}\sqrt{{\left(48\right)}^2+{\left(14\right)}^2}\text{=}\sqrt{2304+196}\text{=}\sqrt{2500}\text{=50 cm} \end{gathered}$$

So, slant height of the cone $=50\text{cm}$

Step-3: curved surface area of the cone

And we know that curved surface area of the cone $=\mathrm{\pi rl}$

After putting the values we get curved surface area $=\frac{22}{7}\times 14\times 50=2200{\text{cm}}^2\left[\because \pi =\frac{22}{7}\right]$

Hence height ,slant height and curve surface area of the cone are $48cm,50cm\text{and}2200c{m}^2$respectively.