Question

From a circle of radius $15$ cm, a sector with ${216}^o$ angle is cut out and its bounding radii are bent so as to form a cone. Then its volume is:

• A. $1081.3{cm}^3$
• B. $1071{cm}^3$
• C. $1018{cm}^3$
• D. None of these

Answer 1

Answer: (C)

$1018{cm}^3$

Given, radius of the circle $\left(R\right)=15$ cm
So, when the sector is cut and its bounding radii is bent to form a cone we have, slant height of the cone, $l=R=15$ cm
Now, let $r$ and $h$ be the radius and height of the cone formed.
Again, we know that in a circle of radius $R$, an arc of length $x$ subtends an angle of $\theta$ radians , then $x=R$ $\theta$
Here, the arc length will be equal to the perimeter of the base circle of the cone.
$\Rightarrow x=2\pi r$
$\Rightarrow 2\pi r=R\theta \Rightarrow \frac{r}{R}=\frac{\theta }{2\pi }\Rightarrow \frac{r}{15}=\frac{216}{360}[Given,\theta ={216}^{\circ }]\Rightarrow r=9$ cm
So, radius of the cone $=9$.
Now, height of the cone can be find out by using pythagoras theorem as:
$h=l-r$
$=15-9=225-81=144$
$\Rightarrow$ $h=12$ cm
Now, volume of the cone $=\frac{1}{3}\pi r^{2}h$

$=\frac{1}{3}\times \frac{22}{7}\times 9^2\times 12$

$=1018.28{cm}^3$
Hence, volume of the cone is approximately $1018$ ${cm}^3$.