A cylinder is closed at both ends by 40 , cm | vedclass

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(B) $1$. The article consists of a cylinder and two cones at its ends.$2$. Radius of cylinder $(r)$ = $30 \, cm$.$3$. Height of each cone $(h_{cone})$

Vedclass · Accepted Answer

$28260$

Vedclass · Answer

(B) $1$. The article consists of a cylinder and two cones at its ends.$2$. Radius of cylinder $(r)$ = $30 \, cm$.$3$. Height of each cone $(h_{cone})$ = $40 \, cm$.$4$. Total height of the article = $180 \, cm$.$5$. Height of the cylinder $(h_{cyl})$ = $180 - (40 + 40) = 100 \, cm$.$6$. Slant height of the cone $(l)$ = $\sqrt{r^2 + h_{cone}^2} = \sqrt{30^2 + 40^2} = \sqrt{900 + 1600} = \sqrt{2500} = 50 \, cm$.$7$. Total Surface Area = (Curved Surface Area of Cylinder) + $2$ $	imes$ (Curved Surface Area of Cone).$8$. Total Surface Area = $(2 \pi r h_{cyl}) + 2 	imes (\pi r l) = 2 \pi r (h_{cyl} + l)$.$9$. Total Surface Area = $2 	imes 3.14 	imes 30 	imes (100 + 50) = 188.4 	imes 150 = 28260 \, cm^2$.

$A$ cylinder is closed at both ends by $40 \, cm$ high cones. The radius of the cylinder is $30 \, cm$ and the total height of the article is $180 \, cm$. Find the total surface area of the article. $(\pi = 3.14)$ (in $cm^2$)

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