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प्रश्न
The interior of a building is in the form of a right circular cylinder of diameter 4.2 m and height 4 m surmounted by a cone of same diameter.
The height of the cone is 2.8 m. Find the outer surface area of the building.
उत्तर
We have,
Radius of the cylinder = Radius of the cone `= "r" = 4.2/2 = 2.1 "m",`
`"Height of the cylinder", "H" = 4 "m"` and
Height of the cone, h = 2.8 m
Also,
The slant height of the cone, `"l" =sqrt("r"^2+"h"^2)`
`=sqrt(2.1^2 + 2.8^2)`
`=sqrt(4.41+7.84)`
`=sqrt(12.25)`
`=3.5 "m"`
Now,
The outer surface area of the building = CSA 0f the cylinder + CSA of the cone
`=2pi"rH" + pi"rl"`
`=pir(2"H"+"l")`
`=22/7xx2.1xx(2xx4+3.5)`
`=6.6xx11.5`
= 75.9 m2
So, the outer surface area of the building is 75.9 m2 .
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