Each edge of a cube is increased by 50 %. The | vedclass

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(A) Let the initial edge of the cube be $x 	ext{ cm}$.The initial surface area of the cube is $S_1 = 6x^2$.The new edge after a $50 \%$ increase is $

Vedclass · Accepted Answer

$125$

Vedclass · Answer

(A) Let the initial edge of the cube be $x 	ext{ cm}$.The initial surface area of the cube is $S_1 = 6x^2$.The new edge after a $50 \%$ increase is $x + 0.5x = 1.5x = \frac{3x}{2} 	ext{ cm}$.The new surface area is $S_2 = 6 \left( \frac{3x}{2} ight)^2 = 6 \left( \frac{9x^2}{4} ight) = \frac{54x^2}{4} = 13.5x^2$.The increase in surface area is $S_2 - S_1 = 13.5x^2 - 6x^2 = 7.5x^2$.The percentage increase is $\left( \frac{	ext{Increase}}{	ext{Initial Area}} ight) 	imes 100 = \left( \frac{7.5x^2}{6x^2} ight) 	imes 100 = 1.25 	imes 100 = 125 \%$.

Each edge of a cube is increased by $50 \%$. Then,the percentage increase in its surface area is (in $\%$)?

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