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Question
An oil funnel made of tin sheet consists of a 10 cm long cylindrical portion attached to a frustum of a cone. If the total height is 22 cm, diameter of the cylindrical portion is 8 cm and the diameter of the top of the funnel is 18 cm, then find the area of the tin sheet required to make the funnel.
Solution
We have,
Height of the cylindrical portion, h = 10 cm,
height of the frustum of cone portion, H = 22 - 10 = 12 cm,
Radius of the cylindrical portion = Radius of smaller end of frustum portion,
Radius of larger end of frustum portion, R = 18/2 = 9 cm
Also, the slant height of the frustum, `l = sqrt(("R - r")^2 + "H"^2)`
`=sqrt((9 - 4)^2+12^2`
`=sqrt(5^2+12^2)`
`=sqrt(25+144`
`=sqrt(169)`
`=13 "cm"`
Now,
The area of the tin sheet required = CSA of frustum of cone + CSAA of cylinder
= π (R + r) l + 2πrh
`= 22/7xx(9+4)x13+2xx22/7xx4xx10`
`=22/7xx13xx13+22/7xx80`
`= 22/7xx(169+80)`
`=22/7xx249`
≈ 782.57 cm2
So, the area of the tin sheet required to make the funnel is 782.57 cm2.
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