[44%] The percentage increase in the area of a rectangle, if each of its sides is increased by 20% is:

For Competitive Exams

The percentage increase in the area of a rectangle, if each of its sides is increased by 20% is:

40%

42%

44%

46%

Explanation:

Let original length = $x$ metres and original breadth = $y$ metres.

Original area = $(xy)$ m².

New length =$ \left(\dfrac{120}{100} x\right) $m=$ \left(\dfrac{6}{5} x\right) $m.

New breadth =$ \left(\dfrac{120}{100} y\right) $m=$ \left(\dfrac{6}{5} y\right) $m.

New Area =$ \left(\dfrac{6}{5} x \times\dfrac{6}{5} y\right) $m²=$ \left(\dfrac{36}{25} xy\right) $m².

The difference between the original area = $xy$ and new-area 36/25 $xy$ is

= (36/25)$x$$y$ - $x$$y$

= $x$$y$(36/25 - 1)

= $x$$y$(11/25) or (11/25)$x$$y$

$\therefore$ Increase % =$ \left(\dfrac{11}{25} xy \times\dfrac{1}{25} \times 100\right) $%= 44%.

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