A tent is in the shape of a right circular cy | vedclass

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(C) The height of the cylindrical part is $h_1 = 3\, m$.The total height of the tent is $H = 13.5\, m$.The height of the conical part is $h_2 = H - h_

Vedclass · Accepted Answer

$2068$

Vedclass · Answer

(C) The height of the cylindrical part is $h_1 = 3\, m$.The total height of the tent is $H = 13.5\, m$.The height of the conical part is $h_2 = H - h_1 = 13.5 - 3 = 10.5\, m$.The radius of the base is $r = 14\, m$.Slant height of the cone $(l) = \sqrt{r^2 + h_2^2} = \sqrt{14^2 + 10.5^2} = \sqrt{196 + 110.25} = \sqrt{306.25} = 17.5\, m$.Curved surface area of the cone $= \pi r l = \frac{22}{7} 	imes 14 	imes 17.5 = 22 	imes 2 	imes 17.5 = 770\, m^2$.Curved surface area of the cylinder $= 2 \pi r h_1 = 2 	imes \frac{22}{7} 	imes 14 	imes 3 = 2 	imes 22 	imes 2 	imes 3 = 264\, m^2$.Total surface area to be painted $= 770 + 264 = 1034\, m^2$.Total cost of painting $= 1034 	imes 2 = ₹ 2068$.

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