A shuttle cock 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,the height of the entire shuttle cock is 7 cm. Find the external surface area.

CSA of frustum = $$\pi\ \left(R+r\right)L$$

CSA of hemisphere = $$2\pi\ r^2$$

Let L be the slant height of the frustum,

L = $$\sqrt{H^2+\left(R-r\right)^2\ }$$

=$$\sqrt{6^2+\left(2.5-1\right)^2\ }$$

=$$\sqrt{36+2.25}=6.18$$ cm

External surface area = CSA of frustum + CSA of hemisphere

$$\pi\ \left(R+r\right)L + 2\pi\ r^2$$

$$\pi\ \left[\left(2.5+1\right)6.18+2\left(1\right)^2\right]\ cm^2$$

$$\frac{22}{7}\left(3.5\times\ 6.18+2\right)$$

$$\frac{22}{7}\times\ 23.63$$

$$74.26\ \ cm2$$