Question

A shuttlecock used for playing badminton has the shape of a frustum of a cone mounted on a hemisphere as shown in the figure. The external diameters of the frustum are $5cm$ and $2cm$, the height of the entire shuttlecock is $7cm$. Find its external surface area.

Answer 1

We have
$r_1=$ Radius of the lower end of the frustum $=1cm$
$r_2=$ Radius of the upper end of the frustum $=2.5cm$
$h=$ Height of the frustum $=6cm$
$l=$ Slant height of the frustum

$\Rightarrow l=\sqrt{{(r_1-r_2)}^2+h^2}$

$=\sqrt{h^2+{(r_1-r_2)}^2}$

$=\sqrt{36+{(2.5-1)}^2}$

$=\sqrt{38.25}$

$=6.18cm$

External surface area of shuttle cock

$=$ curved surface area of the frustum + surface area of hemisphere

$=\pi (r_1-r_2)l+2\pi r_1^2$

$=\{\pi (1+2.5)\times 6.18+2\times \pi \times 1^2\}{cm}^2$

$=\{\frac{22}{7}\times 3.5\times 6.18+2\times \frac{22}{7}\}{cm}^2$

$=74.26{cm}^2$