Question

A metallic sector of central angle 240${}^o$ and radius $15$ cm is rolled up to form a cone.
What is the volume of the cone so formed?

Answer 1

${240}^o$ is $\frac{240}{360}=\frac{2}{3}$ part of the central angle of the big circle.
So the radius of the cone is $\frac{2}{3}$ part of the radius of the big circle,
Radius of cone $15\times \frac{2}{3}=10cm$
When cone is made using the sector, its slant height = 15 cm
$\Rightarrow \text{Height of the cone}=\sqrt{{15}^2-{10}^2}=\sqrt{225-100}=\sqrt{125}=5\sqrt{5}cm$
$\text{Volume of the cone}=\frac{1}{3}\pi r^{2}h=\frac{1}{3}\times \pi \times 10\times 10\times 5\sqrt{5}=\frac{500\sqrt{5}}{3}\pi c{m}^3$