A solid cylinder has a base radius of 14, cm | vedclass

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(B) The total surface area of the remaining part consists of: $1$. The curved surface area of the original cylinder: $2\pi Rh = 2 	imes \frac{22}{7}

Vedclass · Accepted Answer

$3432$

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(B) The total surface area of the remaining part consists of: $1$. The curved surface area of the original cylinder: $2\pi Rh = 2 	imes \frac{22}{7} 	imes 14 	imes 15 = 1320\, cm^2$. $2$. The area of the two bases of the original cylinder minus the area of the two circular cuts: $2 	imes (\pi R^2 - \pi r^2) = 2 	imes \pi 	imes (14^2 - 7^2) = 2 	imes \frac{22}{7} 	imes (196 - 49) = 2 	imes \frac{22}{7} 	imes 147 = 924\, cm^2$. $3$. The curved surface area of the two small cylinders cut out: $2 	imes (2\pi rh) = 2 	imes (2 	imes \frac{22}{7} 	imes 7 	imes 5) = 880\, cm^2$. $4$. The area of the inner circular base of the two small cylinders: $2 	imes (\pi r^2) = 2 	imes \frac{22}{7} 	imes 7^2 = 308\, cm^2$. Total surface area = $1320 + 924 + 880 + 308 = 3432\, cm^2$.

$A$ solid cylinder has a base radius of $14\, cm$ and a height of $15\, cm$. Identical small cylinders are cut from each base as shown in the figure. The height of each small cylinder is $5\, cm$ and the radius is $7\, cm$. What is the total surface area (in $cm^2$) of the remaining part?

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