A sector of radius 10.5 cm with the central angle $$120^\circ$$ is folded to form a cone by joining the two bounding radii of the sector. What is the volume (in $$cm^3$$) of the cone so formed?

When a sector of a circle is folded to form a cone.

The slant height of the cone = radius of the circle = 10.5cm

The base of the cone forms a sector of circle equal in length to the length of the arc.

Perimeter of the sector of the circle = length of base of cone

2 $$\times \pi \times r \times \frac{angle}{360} = 2 \times \pi \times r1$$

(Let the radius of cone r1)

r1 = $$\frac{10.5}{3}$$ = 3.5 cm

Height of cone = h 

by pythagoras theorem-

$$h^2 = (10.5)^2 - (3.5)^2$$

$$h^2 = 110.25 - 12.25$$

h = $$\sqrt{98}$$

Volume of cone = 1/3 $$\times \pi \times r^2 \times h$$ 

= 1/3 $$\times \pi \times (3.5)^2 \times \sqrt{98} = \frac{\pi \times 85.75 \sqrt{2}}{3} = \frac{343\sqrt{2}}{12}\pi$$