If the side of a cube is increased by 5 %,the | vedclass

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(D) Let the side of the cube be $x$. The surface area $S$ of the cube is given by $S = 6x^2$. When the side $x$ is increased by $5 \%$,the new side be

Vedclass · Accepted Answer

$10.25$

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(D) Let the side of the cube be $x$. The surface area $S$ of the cube is given by $S = 6x^2$. When the side $x$ is increased by $5 \%$,the new side becomes $x' = x + 0.05x = 1.05x$. The new surface area $S'$ is given by $S' = 6(1.05x)^2$. $S' = 6(1.1025x^2) = 1.1025(6x^2) = 1.1025S$. The percentage increase in surface area is $\frac{S' - S}{S} 	imes 100 \%$. $= \frac{1.1025S - S}{S} 	imes 100 \% = 0.1025 	imes 100 \% = 10.25 \%$.

If the side of a cube is increased by $5 \%$,then the surface area of a cube is increased by (in $\%$)

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