The length and breadth of a square are increa | vedclass

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(B) Let the side of the square be $s$. The area of the square is $s^2$.When the length is increased by $30 \%$,the new length becomes $L = s(1 + 0.30)

Vedclass · Accepted Answer

$56 \%$

Vedclass · Answer

(B) Let the side of the square be $s$. The area of the square is $s^2$.When the length is increased by $30 \%$,the new length becomes $L = s(1 + 0.30) = 1.3s$.When the breadth is increased by $20 \%$,the new breadth becomes $B = s(1 + 0.20) = 1.2s$.The area of the rectangle formed is $L 	imes B = (1.3s) 	imes (1.2s) = 1.56s^2$.The increase in area is $1.56s^2 - s^2 = 0.56s^2$.The percentage increase in area is $\frac{0.56s^2}{s^2} 	imes 100 = 56 \%$.Alternatively,using the net percentage change formula: $	ext{Net } \% = x + y + \frac{xy}{100} = 30 + 20 + \frac{30 	imes 20}{100} = 50 + 6 = 56 \%$.

The length and breadth of a square are increased by $30 \%$ and $20 \%,$ respectively. The area of the rectangle so formed exceeds the area of the square by:

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