from a solid cylinder whose high is 2.4cm and diameter 1.4cm a conical conity of the same height and same diameter is hallowed out. fiend the total surface area of the remaining solid to the nearest cm.
Answers
Answer:
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