Question

A solid is in the form of a cylinder surmounted by a cone of the same radius. If the radius of the base and the height of the cone are 'r' and 'h' cm respectively and the total height of the solid is 3h, prove that the volume of the solid is 1/3πhr^2.

Answer 1

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Volume of the solid=Volume of cone+Volume of cylinder

Volume of cone=1/3πr^2h

=1/3x22/7x7x7x9

=462cm^3

Height of cylinder=Total height of solid-height of cone

=30-9

=21cm

Volume of cylinder=πr^2h

=22/7x7x7x21

=3234cm^3

Therefore, Volume of solid=462+3234

=3696cm^3

Answer 2

Answer:

Volume of the solid is 7/3πhr^2.

Step-by-step explanation:

radius of the base and the height of the cone are 'r' and 'h' cm respectively

Volume of the cone is $V_{cone}= \frac{\pi }{3} r^2h$

Radius of cylinder is r

Total height of the solid is 3h, we know cone is of h. hence height of the cylinder is 3h -h = 2h

Volume of the cylinder is $V_{cylinder}=2\pi r^2 h$

Cone is  surmounted on the cylinder, so volume will be added for total volume

$V= V_{cone}+V_{cylinder}\\V= \frac{\pi }{3} r^2h+2\pi r^2h\\V= \pi r^2h(\frac{1}{3} +2)\\V=\pi r^2h(7/3)\\V=\frac{7}{3} \pi r^2h$

Volume of the solid is 7/3πhr^2.