The dome of a temple is in the shape of a con | vedclass

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(C) Given: Radius $(r)$ = $3.5\, m$,Height $(h)$ = $12\, m$.First,find the slant height $(l)$ of the cone using the formula $l = \sqrt{r^2 + h^2}$.$l

Vedclass · Accepted Answer

$4125$

Vedclass · Answer

(C) Given: Radius $(r)$ = $3.5\, m$,Height $(h)$ = $12\, m$.First,find the slant height $(l)$ of the cone using the formula $l = \sqrt{r^2 + h^2}$.$l = \sqrt{(3.5)^2 + (12)^2} = \sqrt{12.25 + 144} = \sqrt{156.25} = 12.5\, m$.The curved surface area $(CSA)$ of the cone is given by $\pi rl$.$CSA = \frac{22}{7} 	imes 3.5 	imes 12.5 = 22 	imes 0.5 	imes 12.5 = 11 	imes 12.5 = 137.5\, m^2$.The cost of painting is at the rate of ₹ $30/m^2$.Total cost = $137.5 	imes 30 = ₹ 4125$.

The dome of a temple is in the shape of a cone with radius $3.5\, m$ and height $12\, m$. Find the cost of painting its outer surface at the rate of ₹ $30/m^2$.

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