A cylinder is closed at both ends by 12 , cm | vedclass

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(B) $1$. Identify the dimensions: Radius $(r)$ = $5 \, cm$,Height of each cone $(h_{cone})$ = $12 \, cm$,Total height $(H)$ = $41 \, cm$.$2$. Calculat

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$942$

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(B) $1$. Identify the dimensions: Radius $(r)$ = $5 \, cm$,Height of each cone $(h_{cone})$ = $12 \, cm$,Total height $(H)$ = $41 \, cm$.$2$. Calculate the height of the cylinder $(h_{cyl})$: $h_{cyl} = H - 2 	imes h_{cone} = 41 - 2(12) = 41 - 24 = 17 \, cm$.$3$. Calculate the slant height $(l)$ of the cone: $l = \sqrt{r^2 + h_{cone}^2} = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13 \, cm$.$4$. The total surface area of the article consists of the curved surface area of the cylinder and the curved surface areas of the two cones.$5$. Total Surface Area = $2 \pi r h_{cyl} + 2 	imes (\pi r l) = 2 \pi r (h_{cyl} + l)$.$6$. Substitute the values: $2 	imes 3.14 	imes 5 	imes (17 + 13) = 31.4 	imes 30 = 942 \, cm^2$.

$A$ cylinder is closed at both ends by $12 \, cm$ high cones. The radius of the cylinder is $5 \, cm$ and the total height of the article is $41 \, cm$. Find the total surface area of the article. $(\pi = 3.14)$ (in $cm^2$)

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