Question

from a solid cylinder whose height is 2.4 CM and diameter is 1.4 CM a conical cavity of the same height and same diameter is hollowed out find the total surface area of remaining solid to the nearest cm²​

Answer 1

Given:

13.6KB/s O a l l 76

From a solid cylinder ..

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Height (h) of the conical part = Height (h) of the cylindrical part = 2.4 cm

Diameter of the cylindrical part = 1.4 cm

Radius =

Diameter

2

Radius(r) of the cylindrical part = 0.7 cm

Slant height (1) of conical part = Vr? + h? =

V0.72 + 2.42 = V

V0.49 + 5.76 = V6.25 = 2.5

Total surface area of the remaining solid CSA of

cylindrical part + CSA of conical part + Area of cylindrical base

= 2trh + Trl + Tr2

22 = 2 x = x 0.7 x 22

2.4 +2.2 + 2.2 x 0.7 × 22x 0.7 x 2.5 + x 0.7 x 0.7

= 4.4 x 2.4 +2.2 x

= 10.56 + 5.50

1.54 = 17.60 cm

The total surface area of the remaining solid to

the nearest cm? is 18 cm²

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