from a cylinder if a conical cavity is removed what will be the formula
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Given that height ( h ) of the conical part = Height ( h ) of the cylindrical part = 2.8 cm
Diameter of the cylindrical part = 4.2 cm
Therefore, radius ( r ) of the cylindrical part = 2.1 cm
CSA of cylindrical part = 2πrh
= 2 x (22/7) x 2.1x 2.8
= 36.96 cm2
CSA of conical part = πrl , where l = √(h2 + r2)
l = √(7.84 + 4.41)
l = 3.5
CSA of conical part = (22/7)x 2.1 x 3.5
= 23.1 cm2
Area of cylindrical base = πr2 = (22/7) x 2.1x2.1
= 13.86 cm2
Total surface area of the remaining solid will be
= CSA of cylindrical part + CSA 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 to the nearest 74 cm2.
Diameter of the cylindrical part = 4.2 cm
Therefore, radius ( r ) of the cylindrical part = 2.1 cm
CSA of cylindrical part = 2πrh
= 2 x (22/7) x 2.1x 2.8
= 36.96 cm2
CSA of conical part = πrl , where l = √(h2 + r2)
l = √(7.84 + 4.41)
l = 3.5
CSA of conical part = (22/7)x 2.1 x 3.5
= 23.1 cm2
Area of cylindrical base = πr2 = (22/7) x 2.1x2.1
= 13.86 cm2
Total surface area of the remaining solid will be
= CSA of cylindrical part + CSA 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 to the nearest 74 cm2.
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