A cylinder is closed at both the ends by 28 , | vedclass

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Question

(B) The article consists of a cylinder with two cones attached at its ends.Radius of cylinder $(r)$ = $21 \, cm$.Height of cylinder $(h)$ = $50 \, cm$

Vedclass · Accepted Answer

$11220$

Vedclass · Answer

(B) The article consists of a cylinder with two cones attached at its ends.Radius of cylinder $(r)$ = $21 \, cm$.Height of cylinder $(h)$ = $50 \, cm$.Height of cone $(H)$ = $28 \, cm$.Slant height of cone $(l)$ = $\sqrt{r^2 + H^2} = \sqrt{21^2 + 28^2} = \sqrt{441 + 784} = \sqrt{1225} = 35 \, cm$.The total surface area of the article = (Curved surface area of cylinder) + $2 	imes$ (Curved surface area of cone).Curved surface area of cylinder = $2 \pi r h = 2 	imes \frac{22}{7} 	imes 21 	imes 50 = 44 	imes 3 	imes 50 = 6600 \, cm^2$.Curved surface area of one cone = $\pi r l = \frac{22}{7} 	imes 21 	imes 35 = 22 	imes 3 	imes 35 = 2310 \, cm^2$.Total surface area = $6600 + 2 	imes 2310 = 6600 + 4620 = 11220 \, cm^2$.

$A$ cylinder is closed at both the ends by $28 \, cm$ high cones. The radius and the height of the cylinder are $21 \, cm$ and $50 \, cm$ respectively. Find the total surface area of the article in $cm^2$.

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