Question

plzz answer it ...
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From a solid cylinder whose height is 2.4 cm and diameter 1.4 cm, a conical cavity of the
same height and same diameter is hollowed out. Find the total surface area of the
remaining solid to the nearest cm​

Answer 1

Step-by-step explanation:

The total surface area of the remaining solid will be 18 cm²

Step-by-step explanation:

The outer surface area of the cylinder = 2πrh

= 2 x 3.14 x 0.7 x 2.4

= 10.55 cm²

slant height of the cone, L = √(h² + r²)

=√(2.4² + 0.7²)

= √6.25

= 2.5 cm

hence outer surface area of the cone = πrL

= 3.14 x 0.7 x 2.5

= 5.5 cm²

This outer surface area of the cone is equal to the inner surface area of the hollow portion of the cylinder left.

surface area of the cylindrical base = πr²

= 3.14 x 0.7²

= 1.54 cm²

Hence total surface area of the remaining solid = The outer surface area of the cylinder + inner surface area of the hollow portion of the cylinder left + surface area of the cylindrical base

= 10.55 + 5.5 + 1.54

= 17.59 cm²

= 18 cm² (rounded off to nearest cm²)

Hence the total surface area of the remaining solid will be 18 cm²

Answer 2

Step-by-step explanation:

thank u ...............