The length of a given rectangle is increased by 20% and the breadth is decreased by 20%. Then the resulting area is

Given, length is increased by $20\%$

$l=l+l\times \frac{20}{100}$

$=l+\frac{l}{5}$

$l=\frac{6l}{5}$

Breadth is decreased by$20\%$

$b=b-b\times \frac{20}{100}$

$=b-\frac{b}{5}$

$b=\frac{4b}{5}$

Area of new rectangle $=\frac{6l}{5}\times \frac{4b}{5}$

$=\frac{24}{25}lb$

Decreased area$=lb-\frac{24}{25}lb$

$=\frac{lb}{25}$

Percentage of decreased area$=\frac{\frac{lb}{25}}{lb}\times 100$

$=4\%$

Decreases by $4\%$

(or)

$a=+20\%$

$b=-20\%$

Percentage of increased Area = $\left(a+b+\frac{ab}{100}\right)\%$

$=\left[+20-20+\frac{+20\times -20}{100}\right]\%$

=$-4\%$

$\therefore$  Area is decreases by $4\%$