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Question
A tent is of the shape of a right circular cylinder upto a height of 3 metres and then becomes a right circular cone with a maximum height of 13.5 metres above the ground. Calculate the cost of painting the inner side of the tent at the rate of Rs 2 per square metre, if the radius of the base is 14 metres.
Solution
Let r m be the radius of cylindrical base of cylinder of height by m
r = 14 m and h1 = 3m
Curved surface area of cylinder
`=2pirh_1m^2`
`=25 22/7 xx 14 xx 3m^2`
`=264 m^2`
The radius of cylindrical box of cylinder is also equal to the radius of right circular cons.
Let h2 be the height of cone and l be the slant height of cone
`r = 14m and h_2 = (13.5 - 3)`
`=10.5 m`
`l^2 = r^2 + h_2^2`
`l^2 = (14)^2 + (10.5)^2`
`l^2 =(14)^2 + (10.5)^2`
`l = sqrt(196 + 110.25)`
`sqrt(306.25 ) = 17.5 m`
Curved surface area of the cone
`pirl`
`= 22/7 xx 14 xx 17.5`
Curved surface of area of cone
`= pirl`
`= 22/7 xx 14 17.5`
`=770 m^2`
Therefore,
Total area of tent which is to be painted
= curved surface area of cylinder + curved surface area of cone
`=(264+770)m^2`
`=1034 m^2`
Now cost of painting 1 m2 of inner side of tent = Rs. 2
Cost of painting 1034 m2 inner side of tent
`= 2 xx 1034`
`="Rs. 2068"`
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