A cylinder with radius 6 , cm is closed at on | vedclass

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Question

(C) The article consists of a cylinder and a cone joined together.Given: Radius of cylinder and cone $(r) = 6 \, cm$,Height of cone $(h_{cone}) = 8 \,

Vedclass · Accepted Answer

$640.56$

Vedclass · Answer

(C) The article consists of a cylinder and a cone joined together.Given: Radius of cylinder and cone $(r) = 6 \, cm$,Height of cone $(h_{cone}) = 8 \, cm$,Total height of article $= 20 \, cm$.Height of cylinder $(h_{cyl}) = 20 \, cm - 8 \, cm = 12 \, cm$.Slant height of cone $(l) = \sqrt{r^2 + h_{cone}^2} = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10 \, cm$.Total Surface Area = (Curved Surface Area of Cylinder) + (Curved Surface Area of Cone) + (Area of base of cylinder).Total Surface Area = $(2 \pi r h_{cyl}) + (\pi r l) + (\pi r^2)$.Total Surface Area = $\pi r (2 h_{cyl} + l + r) = 3.14 	imes 6 	imes (2 	imes 12 + 10 + 6)$.Total Surface Area = $18.84 	imes (24 + 10 + 6) = 18.84 	imes 40 = 753.60 \, cm^2$.Wait,re-evaluating: The article is open at the other end,so the base area is not included.Surface Area = (Curved Surface Area of Cylinder) + (Curved Surface Area of Cone) = $2 \pi r h_{cyl} + \pi r l$.Surface Area = $\pi r (2 h_{cyl} + l) = 3.14 	imes 6 	imes (2 	imes 12 + 10) = 18.84 	imes (24 + 10) = 18.84 	imes 34 = 640.56 \, cm^2$.

$A$ cylinder with radius $6 \, cm$ is closed at one end by a $8 \, cm$ high cone and is open at the other end. The total height of the article is $20 \, cm$. Find the total surface area of the article. $(\pi = 3.14)$ (in $cm^2$)

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