If the side of a square is increased by 30 %, | vedclass

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(B) Let the original side of the square be $s$.Original area $A_1 = s^2$.If the side is increased by $30 \%$,the new side $s' = s + 0.30s = 1.3s$.New

Vedclass · Accepted Answer

$69 \%$

Vedclass · Answer

(B) Let the original side of the square be $s$.Original area $A_1 = s^2$.If the side is increased by $30 \%$,the new side $s' = s + 0.30s = 1.3s$.New area $A_2 = (1.3s)^2 = 1.69s^2$.The increase in area is $A_2 - A_1 = 1.69s^2 - s^2 = 0.69s^2$.Percentage increase in area $= \left( \frac{0.69s^2}{s^2} ight) 	imes 100 = 69 \%$.Alternatively,using the net percentage change formula for two successive changes: $x + y + \frac{xy}{100}$,where $x = 30$ and $y = 30$.Net change $= 30 + 30 + \frac{30 	imes 30}{100} = 60 + 9 = 69 \%.$

If the side of a square is increased by $30 \%,$ its area is increased by:

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