A shuttlecock used for playing badminton has the shape of a frustum of the frustum are 5 cm and 2 cm, and the height of the entire shuttlecock is 7 cm. Find its external surface area.

Explanation

Here, `R=5/2cm, r = 1 cm` and
`h=(7-1) cm = 6`
Let l be the slant height of the frustum of the cone.
`:." "l^(2)=h^(2)+(R+r)^(2)={6^(2)+(5/2-1)^(2)}cm^(2)`
`=(36+9/4)cm^(2)=153/4cm^(2)`
`rArr" "l=sqrt(153)/2cm=(12.36)/2cm=6.17 cm.`
External surface area of the shuttlecock
= (its lateral surface area + area of a the base)
`=[pi(R+r)+2pi r^(2)']" sq units"`
`={22/7xx6.18+44/7)cm^(2)=(67.98+6.28)cm^(2)=74.26 cm^(2)`.
`=(11xx6.18+44/7)cm^(2)=(67.98+6.28)cm^(2)=72.26cm^(2)`.