From a circle of radius 15 cm, a sector central angle 216° is cut and…

Here, radius of the circle R = 15 cm

Where the sector is cut and its bounding radii is bent to form a cone.

Slant height of the cone, l = R = 15 cm

Let ‘r’ and ‘h’ be the radius and height of the cone respectively.

Again, we know that in a circle of radius ‘R’, an arc of length ‘X’ subtends an angle of ‘θ’ radians.

Then x = Rθ

$⇒2πr=Rθ⇒\frac{r}{R}=\frac{θ}{2π}$

$⇒\frac{r}{15}=\frac{216}{360}$ ⇒ r = 9 cm

Now, height of the cone can be calculated as

$h^2=l^2-r^2$

$h^2=15^2-9^2=225-51=144$

h = 12 cm

∴ Volume of the cone, $V = \frac{1}{3}πr^2h$

$=\frac{1}{3}×\frac{22}{7}×81×12$

= 1018.28 cm³