Question:
The radii of the top and bottom of a bucket of slant height 45 cm are 28 cm and 7 cm, respectively. Find the curved surface area of the bucket.
Solution:
Let R and r be the radii of the top and base of the bucket, respectively, and let l be its slant height.
Then, curved surface area of the bucket
= Curved surface area of the frustum of the cone
$=\pi l(R+r)$
$=\frac{22}{7} \times 45 \times(28+7) \mathrm{cm}^{2}$
$=\left(\frac{22}{7} \times 45 \times 35\right) \mathrm{cm}^{2}$
$=4950 \mathrm{~cm}^{2}$