A shuttle cock used for playing badminton has the shape of a frustum of a cone mounted on a hemisphere - Maths - Surface Areas and Volumes

Question

A shuttle cock used for playing badminton has the shape of a
frustum of a cone mounted on a hemisphere (see figure) The
diameters of the ends of the frustum are 5 cm and 2 cm , the height
of the entire shuttle cock is 7 cm Find the external surface area
(Take Pie=22/7)

Aakash Sharma · Accepted Answer

Given: r1 = 1 cm, r2 = 2 5 cm, h = 6 cm $l = \sqrt{h^{2} + (r_{2} - r_{1})^{2}} = \sqrt{{(6)}^{2} + (2 5 - 1)^{2}} cm = \sqrt{36 + 2 25} cm = \sqrt{38 25} cm = 6 18 cm$ So, External surface area = C S A of frustum + C S A of hemisphere = Ï€(r1 + r2) l + 2Ï€r12 = Ï€ [ (1 + 2 5) 6 18 + 2 Ã— 12 ] cm2 $= \frac{22}{7} [3 5 \times 6 18 + 2] {cm}^{2} = \frac{22}{7} [21 63 + 2] {cm}^{2} = \frac{22}{7} \times 23 63 {cm}^{2} = \frac{519 86}{7} {cm}^{2} = 74 26 {cm}^{2}$

A shuttle cock used for playing badminton has the shape of a frustum of a cone mounted on a hemisphere (see figure). The diameters of the ends of the frustum are 5 cm and 2 cm , the height of the entire shuttle cock is 7 cm. Find the external surface area. (Take Pie=22/7) .