A sector of radius 10.5 cm with the central angle 120° is folded to form a cone by joining the two bounding radii of the sector. What is the volume (in cm³) of the cone so formed?

Solution (By Examveda Team)

$$\eqalign{ & {\text{Length of Arc}} = L \cr & L = 2 \times \frac{{22}}{7} \times \frac{{r \times 120}}{{360}} \cr & L = 2 \times \frac{{22}}{7} \times 10.5 \times \frac{1}{3} \cr & L = 22 \cr} $$

$$\eqalign{ & 2\pi R = 22 \cr & 2 \times \frac{{22}}{7}R = 22 \cr & R = 3.5 \cr & {h^2} + {R^2} = {10.5^2} \cr & {h^2} + {3.5^2} = {10.5^2} \cr & h = 7\sqrt 2 \cr & V = \frac{1}{3}\pi {r^2}h \cr & V = \frac{1}{3}\pi \times 3.5 \times 3.5 \times 7\sqrt 2 \cr & V = \frac{{343\sqrt 2 }}{{12}}\pi \cr} $$