If the area of a square is increased by 44 %, | vedclass

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(D) Let the initial side of the square be $s_1 = 10 	ext{ cm}$.Therefore,the initial area of the square is $A_1 = (s_1)^2 = 10^2 = 100 	ext{ cm}^2$.

Vedclass · Accepted Answer

$20$

Vedclass · Answer

(D) Let the initial side of the square be $s_1 = 10 	ext{ cm}$.Therefore,the initial area of the square is $A_1 = (s_1)^2 = 10^2 = 100 	ext{ cm}^2$.Given that the area increases by $44 \%$,the new area $A_2$ is:$A_2 = A_1 + 44\% 	ext{ of } A_1 = 100 + 44 = 144 	ext{ cm}^2$.Let the new side of the square be $s_2$. Since $A_2 = (s_2)^2$,we have:$s_2 = \sqrt{144} = 12 	ext{ cm}$.The percentage increase in the side is calculated as:$	ext{Percentage increase} = \frac{s_2 - s_1}{s_1} 	imes 100 = \frac{12 - 10}{10} 	imes 100 = \frac{2}{10} 	imes 100 = 20 \%$.Thus,each side increases by $20 \%$.

If the area of a square is increased by $44 \%$,while retaining its shape as a square,by what percentage does each of its sides increase?

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