Question

The radii of the top and bottom of a bucket of slant height 45 cm are 28 cm and 7 cm, respectively. The curved surface area of the bucket is
(a) 4900 cm²
(b) 4950 cm²
(c) 4960 cm²
(d) 4980 cm²

Answer 1

(b) 4950 cm²
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
$$\begin{gathered} =\mathrm{\pi }l\left(R+r\right) \\ =\frac{22}{7}\times 45\times \left(28+7\right){\text{cm}}^2 \\ =\left(\frac{22}{7}\times 45\times 35\right){\text{cm}}^2 \\ =4950{\text{cm}}^2 \end{gathered}$$