The percentage increase in the surface area o | vedclass

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(A) Let the side of the cube be $a$.The initial surface area of the cube is $S_1 = 6a^2$.When the side is doubled,the new side becomes $2a$.The new su

Vedclass · Accepted Answer

$300$

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(A) Let the side of the cube be $a$.The initial surface area of the cube is $S_1 = 6a^2$.When the side is doubled,the new side becomes $2a$.The new surface area of the cube is $S_2 = 6(2a)^2 = 6(4a^2) = 24a^2$.The increase in surface area is $S_2 - S_1 = 24a^2 - 6a^2 = 18a^2$.The percentage increase is calculated as $\frac{	ext{Increase}}{	ext{Original Area}} 	imes 100$.Percentage increase $= \frac{18a^2}{6a^2} 	imes 100 = 3 	imes 100 = 300\%$.

The percentage increase in the surface area of a cube when each side is doubled is:

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