Increase And Decrease By The Same Amount

Relevant wiki: Length and Area

Consider Area of the rectangle,

$\text {A}$ $=$ $\text {lb}$ ..... $(1)$

Now , length is 20% increased & breadth is 20% decreased , therefore new length & new breadth are given by :-

$\text {l'}$ $=$ $\text {l}$ $+$ $\dfrac {l}{5}$ $=$ $\dfrac {6l}{5}$

$\text {b'}$ $=$ $\text {b}$ $-$ $\dfrac {b}{5}$ $=$ $\dfrac {4b}{5}$

Therefore new area will be given by:-

$\text {A'}$ $=$ $\dfrac {6l}{5}•\dfrac {4b}{5}$

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

$=$ $\dfrac {24A}{25}$ ..... $\text{(from equation 1)}$

:-) It can be clearly seen that new area is a proper fraction of the original area , hence $\Rightarrow$

$\text {Area decreases by some amount !}$ ; Moreover there is a decrease of 4% in the area!

$\textit {Thank you.}$ :-)

Suppose A and B are the sides of a rectangle such that A > B.
Then the area C, is AB = C.
If the larger side A is decreased 20% and the smaller side B is increased 20%, the area D of the new rectangle is (.8A)(1.2B)=D.
Using the associative property we see that (.8A)(1.2B) = (.8)(1.2)AB = .96AB,
Thus the area D = .96AB,
and D = .96AB < AB = C, and therefore the area of the original rectangle, C, is greater than the area of the new rectangle, D.

increase decrease can be calculated as :

$A + B + (\frac{A*B}{100}$ )

here -20 +20 +((-20) *20)/100 = - 4% :)