Question

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

Answer 1

Let $r$ be the base radius of the cone.
Angle of the sector, $\theta ={120}^o$
Radius of the sector, $R=21cm$
When the sector is folded into a right circular cone, we have circumference of the base of the cone$=$ Length of the arc
$\Rightarrow 2\pi r=\frac{\theta }{{360}^o}\times 2\pi R$
$\Rightarrow r=\frac{\theta }{{360}^o}\times R$
Thus, the base radius of the cone, $r=\frac{{120}^o}{{360}^o}\times 21=7cm$
Also, the slant height of the cone,
$l=$ Radius of the sector
Thus, $l=R$ $\Rightarrow$ $l=21cm$
Now, the curved surface area of the cone,
$CSA=\pi rl$
$=\frac{22}{7}\times 21=462$
Thus, the curved surface area of the cone is $462sq.cm$