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

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(A) For both the cones:Radius $r = 5 \, cm$ and height $h = 12 \, cm$.For the cylinder:Radius $r = 5 \, cm$.Height of the cylinder $H = 	ext{Total he

Vedclass · Accepted Answer

$942$

Vedclass · Answer

(A) For both the cones:Radius $r = 5 \, cm$ and height $h = 12 \, cm$.For the cylinder:Radius $r = 5 \, cm$.Height of the cylinder $H = 	ext{Total height of the article} - 2 	imes 	ext{Height of cone}$.$H = 41 - 2 	imes 12 = 41 - 24 = 17 \, cm$.Slant height of the cone $l = \sqrt{r^2 + h^2}$.$l = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13 \, cm$.Total Surface Area $(TSA)$ of the combined article:$TSA = 	ext{CSA of the cylinder} + 2 	imes 	ext{CSA of the cone}$.$TSA = 2 \pi r H + 2 	imes (\pi r l) = 2 \pi r (H + l)$.$TSA = 2 	imes 3.14 	imes 5 	imes (17 + 13)$.$TSA = 3.14 	imes 10 	imes 30 = 3.14 	imes 300 = 942 \, cm^2$.Thus,the total surface area of the given article is $942 \, cm^2$.

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

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