In measuring the sides of a rectangle,one sid | vedclass

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(A) Let the original sides of the rectangle be $L$ and $W$. The original area is $A = L 	imes W$.According to the problem,the new sides are $L' = L(1

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$12 \%$ deficit

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(A) Let the original sides of the rectangle be $L$ and $W$. The original area is $A = L 	imes W$.According to the problem,the new sides are $L' = L(1 + 0.10) = 1.1L$ and $W' = W(1 - 0.20) = 0.8W$.The new area is $A' = L' 	imes W' = (1.1L) 	imes (0.8W) = 0.88 	imes (L 	imes W) = 0.88A$.The change in area is $A' - A = 0.88A - A = -0.12A$.The error percentage is $\left( \frac{-0.12A}{A} ight) 	imes 100 = -12 \%$.Alternatively,using the formula for successive percentage change: $	ext{Net Change} = \left( x + y + \frac{xy}{100} ight) \%$.Here,$x = 10$ and $y = -20$. Substituting these values: $	ext{Net Change} = \left( 10 - 20 + \frac{10 	imes (-20)}{100} ight) \% = (-10 - 2) \% = -12 \%$.$A$ negative sign indicates a deficit. Thus,the error is $12 \%$ deficit.

In measuring the sides of a rectangle,one side is taken $10 \%$ in excess and the other $20 \%$ in deficit. Find the error per cent in area calculated from the measurement.

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