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प्रश्न
From a solid cylinder of height 2.8 cm and diameter 4.2 cm, a conical cavity of the same height and same diameter is hollowed out. Find the total surface area of the remaining solid [take π=22/7]
उत्तर
The following figure shows the required cylinder and the conical cavity.
Given:
Height (h) of the conical part = Height (h) of the cylindrical part = 2.8 cm
Diameter of the cylindrical part = Diameter of the conical part = 4.2 cm
∴ Radius (r) of the cylindrical part = Radius (r) of the conical part = 2.1 cm
Slant height (l) of the conical part `=sqrt(r^2+h^2)`
`=sqrt((2.1)^2+(2.8)^2)cm`
`=sqrt(4.41+7.84)cm`
`=sqrt(12.25)cm=3.5 cm`
Total surface area of the remaining solid = Curved surface area of the cylindrical part + Curved surface area of the conical part + Area of the cylindrical base
`=2pirh+pirl+pir^2`
`=(2xx22/7xx2.1xx2.8+22/7xx2.1xx3.5+22/7xx2.1xx2.1) cm^2`
`=(36.96+23.1+13.86)cm^2`
`=73.92 cm^2`
Thus, the total surface area of the remaining solid is 73.92 cm2.
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