Question

The radius $OP$ of the sector $O-PRQ$ is $12cm$ and $\angle POQ={120}^{\circ }.$ A cone is formed such that radii $OP$ and $OQ$ coincide. Find the volume of the cone
($\pi =3.14$ and $\sqrt{2}=1.41$)

Answer 1

REF.Image

Given In $\Delta OPQ$ $\angle POQ={120}^{\circ }$

and [PO=OQ= 12 cm]

if cone is formed using $\Delta OPQ$

From $\Delta OPM$ $cos60=\frac{OM}{OP}=\frac{DM}{12}$

and $MP=12sin60=6\sqrt{3}cm$

now area of $\Delta OPQ=\frac{1}{2}\times 6\times 12\sqrt{3}$

area of $\Delta OPQ=36\sqrt{3}c{m}^2$

now we know that

if cone is formed then

height of cone = OM = 6 cm

and radius r of cone is

$2\pi r$ = perimeter of lower circle = PQ

$2\pi r=12\sqrt{3}$

$[r=\frac{6\sqrt{3}}{\pi }]$

Volume of cone = $\frac{1}{3}\pi r^{2}h$

$\Rightarrow \frac{1}{3}\times \pi \times (\frac{6\sqrt{3}}{\pi }{)}^2\times 6$

volume of cone = $\frac{36\times 3}{{\pi }^2}\times \pi \times 2$

Volume of cone = $6\times \frac{36}{\pi }=68.7c{m}^3$