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 2 . Given that, Height (h) of the conical part = Height (h) of the cylindrical part = 2.4 cm Diameter of the cylindrical part = 1.4 cm Therefore, radius (r) of the cylindrical part = 0.7 cm Total surface area of the remaining solid will be = CSA of cylindrical part + CSA of conical part + Area of cylindrical base The total surface area of the remaining solid to the nearest cm 2 is 18 cm 2 . Why C.S.A. of cone is added instead of subtracting(cause we have to find T.S.A. of remaining part)??? plz explain
Answers
Answer:
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:
Step-by-step explanation:
Height of cylinder = Height of conical cavity, h = 2.4 cm
Diameter, d = 1.4 cm
Radius, r = 1.4/2= 0.7 cm
Slant height, l = √h² + r² = √(2.4)² + (0.7)² = √5.76 + 0.49 = √6.25 = 2.5 cm
TSA of the remaining solid = CSA of cylindrical part + CSA of conical part + Area of cylindrical base
TSA = 2πrh+ πrl+ πr²
= πr [ 2(h) + l + r ]
= 22/7 × 0.7 [ 2 × 2.4 + 2.5 + 0.7 ]
= 17.6 cm²
Hence,
TSA of remaining solid is = 17.6
Hope it helps you
Cheers!!