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# From a solid cylinder of height 2.8 cm and diameter 4.2 cm, a conical cavity of the same height and same diameter is hollowed out. Find the total surface area of the remaining solid.

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This is question was asked in 2014 cbse board exam. This is the important question from the exam point of view. This question from RS Aggarwal book Exercises 17A page number 788, chapter volume and surface area of solid.

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1. Given data,

Height ( h ) of the conical part = Height ( h ) of the cylindrical part = 2.8 cm

Diameter of the cylindrical part = 4.2 cm

∴ radius ( r ) of the cylindrical part = 2.1 cm

curved surface area of cylindrical part = 2πrh

= 2 × (22/7) × 2.1 × 2.8

= 36.96 cm²

curved surface area of conical part = πrl

l = √(h² + r²)

l = √(7.84 + 4.41)

l = 3.5

curved surface area of conical part = (22/7) × 2.1 × 3.5

= 23.1 cm²

Area of cylindrical base = πr²

= (22/7) × 2.1

= 13.86 cm²

Total surface area of the remaining solid will be

= curved surface area of cylindrical part + curved surface area of conical part + Area of cylindrical base

= 36.96 cm² + 23.1 cm² + 13.86 cm²

= 73.92 cm²

∴ The total surface area of the remaining solid is 74 cm²

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