Question

If the length is increased by 10% and breadth is decreased by 10% of a rectangle, then
what percent of area will be increased or decreased?


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Answer 1

Step-by-step explanation:

AND THE ANSWER IS 1%

PLEASE MAKE ME AS BRILLIANT PLEASE

Answer 2

$\bf{\underline{\underline{To\:Find}}}$

The increase or decrease in area if the length of a rectangle is increased by 10% and breadth is decreased by 10%.

$\bf{\underline{\underline{Assumption}}}$

Let the length be x and breadth be y

$\bf{\underline{\underline{Solution}}}$

$\bf{Length\:of\:original\:rectangle\:(L_1) = x}$

$\bf{Breadth\:of\:original\:rectangle\:(B_1) = y}$

$\bf{Area\:of\:original\:rectangle [A_1] = (Length\times Breadth)\:sq.units}$

= $\bf{A_1 = (x\times y)\:sq.units}$

= $\bf{A_1 = xy\:sq.units}$

Now,

$\bf{New\:length\:of\:rectangle\:(L_2) = x + 10\%of\:x}$

= $\bf{L_2 = x + x\times\dfrac{10}{100}}$

= $\bf{L_2 = x + \dfrac{x}{10}}$

= $\bf{L_2 = \dfrac{10x + x}{10}}$

= $\bf{L_2 = \dfrac{11x}{10}}$

$\bf{Breadth\:of\:new\:rectangle (B_2) = y-10\%of\:y}$

= $\bf{B_2 = y - y\times\dfrac{10}{100}}$

= $\bf{B_2 = y - \dfrac{y}{10}}$

= $\bf{B_2 = \dfrac{10y-y}{10}}$

= $\bf{B_2 = \dfrac{9y}{10}}$

$\bf{Area\:of\:new\:rectangle \:(A_2)= \dfrac{11x}{10}\times\dfrac{9y}{10}\:sq.units}$

= $\bf{A_2 = \dfrac{99xy}{100}}$

Now,

$\bf{Decrease\:in\:Area = xy - \dfrac{99xy}{100}}$

= $\bf{Decrease\:in\:area = \dfrac{100xy - 99xy}{100}}$

= $\bf{Decrease\:in\:area = \dfrac{xy}{100}}$

$\bf{\underline{Therefore}}$

$\bf{Decrease\:in\:percentage = \dfrac{A_2}{A_1}\times100}$

= $\bf{Decrease\:in\:percentage = \dfrac{xy}{\dfrac{100}{xy}}\times100}$

= $\bf{Decrease\:in\:percentage = \dfrac{xy}{100}\times\dfrac{1}{xy}\times100}$

= $\bf{Decrease\:in\:percentage = 1\%}$