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The adjoining figure represents a solid consisting of a cylinder surmounted by a cone at one end and a hemisphere at the other end. Given that common radius =3.5 cm the height of the cylinder =6.5 cm and the total height =12.8 cm , claculate the volume of the solid correct to the nearest cm. |
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Answer» Radius of the hemisphere =3.5 cm=`(7)/(2)`cm Height of the cylinder = 6.5 cm =`(13)/(2)`cm Given total height -12.8 cm Height of the cone =(12.8-3.5-6.5)cm =2.8 cm=`(14)/(5)`cm Radius of cylinder =`(7)/(2)` cm=Radius of cone Volume of the hemisphere=`(2)/(3)pi((7)/(2))^(3)cm^(3)` volume of the cylinder =`pi((7)/(2))^(2)((13)/(2))cm^(3)` Volume of the given solid `=[(2)/(3)pixx(7)/(2^(3))+pixx(7)/(2^(2))xx(13)/(2)+(1)/(3)pixx(7)/(2^(2))xx(14)/(5)]cm^(3)` `=pi[(7)/(2^(2)) ((2)/(3))xx(7)/(2^(2))xx(13)/(2)+(1)pixx(7)/(2^(2))xx(14)/(5)]cm^(3)` `=pi[(7)/(2^(2))(2)/(3)xx(7)/(2^(2))xx(13)/(2)+(1)/(3)pixx(7)/(2^(2))xx(14)/(5)]cm^(3)` `=pi(7)/(2^(2))(2)/(3)xx(7)/(2)+(13)/(2)+(1)/(3)xx((14)/(5))cm^(3)` `=(77)/(2)xx(70+195+28)/(30)cm^(3)=(77xx293)/(60)cm^(3)` `=376 cm^(3)` (approx) |
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