For a rectangle,the length and breadth are in | vedclass

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(A) Let the original length be $L$ and the original breadth be $B$. The original area is $A = L 	imes B$.After the increase,the new length $L' = L +

Vedclass · Accepted Answer

$32 \%$

Vedclass · Answer

(A) Let the original length be $L$ and the original breadth be $B$. The original area is $A = L 	imes B$.After the increase,the new length $L' = L + 0.10L = 1.10L$ and the new breadth $B' = B + 0.20B = 1.20B$.The new area $A' = L' 	imes B' = (1.10L) 	imes (1.20B) = 1.32 	imes (L 	imes B) = 1.32A$.The percentage increase in area is given by $\frac{A' - A}{A} 	imes 100 = \frac{1.32A - A}{A} 	imes 100 = 0.32 	imes 100 = 32 \%$.Alternatively,using the net percentage change formula for two successive changes $x$ and $y$:Net $\%$ change $= (x + y + \frac{xy}{100}) \% = (10 + 20 + \frac{10 	imes 20}{100}) \% = (30 + 2) \% = 32 \%$.

For a rectangle,the length and breadth are increased by $10 \%$ and $20 \%$,respectively. The percentage increase in area is:

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