A solid cylinder having a radius of base as 7 | vedclass

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(B) The total surface area $(TSA)$ of the original solid cylinder is given by the formula: $TSA = 2 \pi r(r + h)$.Substituting the values $r = 7 \, cm

Vedclass · Accepted Answer

$25.93$

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(B) The total surface area $(TSA)$ of the original solid cylinder is given by the formula: $TSA = 2 \pi r(r + h)$.Substituting the values $r = 7 \, cm$ and $h = 20 \, cm$:$TSA_{original} = 2 	imes \frac{22}{7} 	imes 7 	imes (7 + 20) = 44 	imes 27 = 1188 \, cm^2$.When the cylinder is bisected along its height,we get two identical cylinders,each with height $h' = \frac{20}{2} = 10 \, cm$.Each new cylinder has a $TSA$ of: $TSA_{new} = 2 \pi r(r + h') = 2 	imes \frac{22}{7} 	imes 7 	imes (7 + 10) = 44 	imes 17 = 748 \, cm^2$.The total surface area of the two new cylinders combined is: $2 	imes 748 = 1496 \, cm^2$.The increase in surface area is: $1496 - 1188 = 308 \, cm^2$.The percentage increase is: $\frac{308}{1188} 	imes 100 \approx 25.93 \%$.

$A$ solid cylinder having a radius of base as $7 \, cm$ and length as $20 \, cm$ is bisected along its height to get two identical cylinders. What will be the percentage increase in the total surface area?

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