Question

A sector of a circle of radius 15 cm has the angle 120º. It is rolled up so that the two bounding radii are joined together to form a cone. The volume of the cone is 

(a) \((250\sqrt2π)\) cm³

(b) \(\big[(500\sqrt2)\fracπ3)\big]\) cm³

(c) \(\big[(250\sqrt2)\fracπ3)\big]\) cm³

(d) \(\big[(1000\sqrt2)\fracπ3)\big]\) cm³

Answer 1

(c) \(\big[(250\sqrt2)\fracπ3)\big]\) cm³

Curved surface area of cone = Area of the sector of circle 

⇒ πrl = πR² x \(\frac{120}{360}\) 

(∵ l = R)

∴ r = 15 x \(\frac{120}{360}\) = 5 cm 

∴ h = \(\sqrt{225 – 25}\) = 10√2 cm 

= Volume of the cone = \(\frac13\) πr²h = \(\frac13\) x π x 25 x 10√2 

=  \(\big[(250\sqrt2)\fracπ3)\big]\) cm³