A shuttlecock used for playing badminton has the shape of a frustum of a Cone mounted on a hemisphere. The external diameters of the frustum are 5 cm and 2 cm, and the height of the entire shuttlecock is 7cm. Find the external surface area.(Ans: 74.26cm2)
Ans: r_{1} = radius of lower end of frustum = 1 cm
r_{2} = radius of upper end = 2.5 cm
l = = 6.18 cm
External surface area of shuttlecock = π (r_{1} + r_{2}) l + 2π r^{2} 1
On substituting we get, = 74.26 cm^{2}