Question

A sector containing an angle of ${120}^{\circ }$ is cut off from a circle of radius 21cm and folded into a cone. Find the curved surface area of the cone. $\left(Take\pi ={\displaystyle \frac{22}{7}}\right)$

Answer 1

Hint: First of all, we will find the radius of base of the cone and then, we will find the slant height of the cone and we will use the formula of curved surface area of cone given by as follows,
$$\mathrm{C}\mathrm{u}\mathrm{r}\mathrm{v}\mathrm{e}\mathrm{d}\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}\mathrm{a}\mathrm{c}\mathrm{e}\mathrm{a}\mathrm{r}\mathrm{e}\mathrm{a}=\pi \mathrm{r}\mathrm{l}$$
Where, r is radius and l is slant height of a cone.
Also, we will use the formula to find the length of the arc given by $R\theta$. Where, R is radius of circle and $\theta$ is angle subtended by the arc at circle.

Complete step-by-step answer:
We have been given a sector containing an angle of ${120}^{\circ }$ is cut off from a circle of radius 21cm and folded into a cone.
Let the base radius of the cone be r and l be the slant height.

Since, the sector is folded into a right circular cone, we have circumference of the base of the cone = length of the arc.
We know that, length of an arc having radius R which subtend an angle $\theta$ at center is equal to $2\pi R\times {\displaystyle \frac{\theta }{{360}^{\circ }}}$
$$\Rightarrow 2\pi r={\displaystyle \frac{\theta }{{360}^{\circ }}}\times 2\pi R$$
Cancelling similar terms on both sides, we get:
$$r={\displaystyle \frac{\theta }{{360}^{\circ }}}\times R$$
Now, we can substitute $\theta ={120}^{\circ }\mathrm{a}\mathrm{n}\mathrm{d}\mathrm{R}=21\mathrm{c}\mathrm{m}.$ So, we will get,
$$\begin{array}{l}\Rightarrow r={\displaystyle \frac{120}{360}}\times 21\\ \Rightarrow r={\displaystyle \frac{1}{3}}\times 21\\ \Rightarrow r=7cm\end{array}$$
Thus, the base radius of cone = 7cm.
Also, the slant height (l) of the cone = Radius of the sector.
Thus, l = R = 21cm.
Now, we know that,
$$\begin{array}{l}\mathrm{C}\mathrm{u}\mathrm{r}\mathrm{v}\mathrm{e}\mathrm{d}\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}\mathrm{a}\mathrm{c}\mathrm{e}\mathrm{a}\mathrm{r}\mathrm{e}\mathrm{a}\mathrm{o}\mathrm{f}\mathrm{a}\mathrm{c}\mathrm{o}\mathrm{n}\mathrm{e}=\pi \mathrm{r}\mathrm{l}\\ \Rightarrow {\displaystyle \frac{22}{7}}\times 21\times 7\\ \Rightarrow 462c{m}^2\end{array}$$

Thus, the curved surface area of the cone is 462 sq.cm

Note: We can also solve this question in less time, if we know that the area of the sector is equal to the area of cone formed by the sector. So, we will find the area of sector using the formula $\pi R^2\times {\displaystyle \frac{\theta }{{360}^{\circ }}}$ , which is also equal to the curved surface area of the cone.