Height of the cone (h) = 24 cm Let r сm be the radius of the base and l cm be the slant height of the cone. Then, l = root under (√r2+ h2 ) = root under (√r2 + 242) = root under (√r2 + 576) Now, Curved surface area = πrl ⇒ 22/7 x r x root under (√r2 + 576) = 550 ⇒ r root under (√r2 + 576) = 550 x 7/22 ⇒ r root under (√r2 + 576) = 175 Squaring both the sides we get r2(r2 + 576) = 30625 (r2)2 + 576r2 - 30625 = 0 Let r2 = x ∴ x2 + 576x - 30625 = 0 x2 + 625x – 49x - 30625 = 0 x(x + 625) - 49(x + 625) = 0 (x + 625)(x - 49) = 0 x + 625 = 0, x - 49 = 0 x = - 625, x = 49 x is not equals to - 625, x = 49 ∴ r2 = 49 ⇒ r = √49 ⇒ r = 7 cm ∴ Volume of the cone = 1/3πr2h = 1/3 x 22/7 x 72 x 24 = 1232 cm3