Question

From a solid right circular cylinder with height 10cm and radius of the base 6 cm, a right circular cone of the same height and base is removed. Find the volume of the remaining solid. ​also find the whole surface area. T.S.A

Answer 1

We are given a solid right circular cylinder from which a right circular cone is removed.

Height of the cylinder = 10 cm = Height of the cone
Radius of the cylinder = 6 cm = Radius of the cone

Volume of the remaining solid = Volume of the cylinder - Volume of the cone

= πr^2h - (πr^2h)/3

$= \frac{22}{7} \times 6 \times 6 \times 10 - \frac{1}{3} \times \frac{22}{7} \times 6 \times 6 \times 10 \\ \\ = 1131.42 - 377.14 \\ \\ = 754.28 \: {cm}^{3}$

(Slant height of cone)^2 = (height of the cone) ^2 + (Radius of the cone) ^2

${l}^{2} = {h}^{2} + {r}^{2} \\ \\ {l}^{2} = {10}^{2} + {6}^{2} \\ \\ {l}^{2} = 100 + 36 \\ \\ {l}^{2} = 136 \\ \\ l = 11.67 \: cm$



The required surface area = Curved surface area of cylinder + area of base of cylinder + curved surface area of cone

=2πrh + πr^2 + πrl

=πr(2h + r + l)

$= \frac{22}{7} \times 6 (2 \times 10 + 6 + 11.67) \\ \\ = \frac{22}{7} \times 6(37.67) \\ \\ = 710.37 \: {cm}^{2}$

Thus, the volume of the remaining solid is 11.67 cubic cms and the total surface area of the solid is 710.37 sq. Cms.