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(D) Radius of the cylinder = radius of the cone = radius of the hemisphere = $r = 14 \, cm$.Height of the cylinder $H = 22 \, cm$.Height of the cone $

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$5368$

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(D) Radius of the cylinder = radius of the cone = radius of the hemisphere = $r = 14 \, cm$.Height of the cylinder $H = 22 \, cm$.Height of the cone $h = 48 \, cm$.Slant height of the cone $l = \sqrt{r^2 + h^2} = \sqrt{14^2 + 48^2} = \sqrt{196 + 2304} = \sqrt{2500} = 50 \, cm$.The total surface area $(TSA)$ of the combined article is the sum of the curved surface areas $(CSA)$ of the cylinder,the cone,and the hemisphere.$TSA = CSA_{	ext{cylinder}} + CSA_{	ext{cone}} + CSA_{	ext{hemisphere}}$$TSA = 2\pi rH + \pi rl + 2\pi r^2 = \pi r(2H + l + 2r)$.Substituting the values: $TSA = \frac{22}{7} 	imes 14 	imes (2 	imes 22 + 50 + 2 	imes 14)$.$TSA = 44 	imes (44 + 50 + 28) = 44 	imes 122 = 5368 \, cm^2$.Thus,the total surface area of the article is $5368 \, cm^2$.

$A$ cylinder with radius $14 \, cm$ is closed at one end by a $48 \, cm$ high cone and at the other end by a hemisphere. If the height of the cylinder is $22 \, cm$,find the total surface area of the article (in $cm^2$).

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