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प्रश्न
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 cm2
[use `pi = 22/7`]
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उत्तर
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
Slant height (l) of conical part =` sqrt(r^2 + h^2)`
`= sqrt((0.7)^2+(2.4)^2) `
`=sqrt(0.49+5.76)`
`=sqrt(6.25)`
= 2.5
Total surface area of the remaining solid will be = CSA of cylindrical part + CSA of conical part + Area of cylindrical base
= 2πrh + πrl + πr2
`= 2xx 22/7xx 0.7xx2.4+22/7xx0.7xx2.5+22/7xx0.7xx0.7`
= `4.4 xx 2.4 + 2.2 xx 2.5 + 2.2 xx 0.7`
= 10.56 + 5.50 +1.54
= 17.60 cm2
The total surface area of the remaining solid to the nearest cm2 is 18 cm2.
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