Question

On a Hemisphere, frustrum of a cone shaped shuttle-cock is used for playing Badminton. The outer radius of the frustrum of cone is $5cm$ and its inner radius is $2cm$. The height of the entire shuttle-cock is $7cm$. Find the outer surface area of the shuttle-cock.

Answer 1

Radius of hemisphere $r=$ smaller radius of frustrum of cone $=2cm$
For the frustrum of cone
$r_1=$ bigger radius $=5cm$
$r_2=$ smaller radius $=2cm$
Height$\left(h\right)=$ total height $-$ radius of hemisphere
$=7-2=5cm$
Now, $l=\sqrt{h^2+{(r_1-r_2)}^2}=\sqrt{5^2+{(5-2)}^2}=\sqrt{25+9}=\sqrt{34}$
$l=5.83cm=$ slant height of the frustrum of a cone.
External surface area of the shuttle cock
$=$ Curved Surface Area of frustum of a cone $+$ Curved Surface Area of hemisphere
$=\pi l(r_1+r_2)+2\pi r^2$
$=\pi [l(r_1+r_2)+2r^2]$
$=\frac{22}{7}\left[5.83\right(5+2)+2(4\left)\right]$
$=\frac{22}{7}[5.83\times 7+8]$
$=\frac{22}{7}\times 48.81=153.40{cm}^2$

Hence, outer surface area is $153.40c{m}^2$