Question

If the bredth of the rectangle is increased by 40% and length is reduced by 30%. what will be the effects to its area?

Answer 1

I suppose an equal sided rectangle *(makes no difference, just easier to show) and denote the length of its sides by “a”

The area of the original rectangle is axa = a square Now do the changes. One side becomes (a+.40) the other (a-.30) The area of the new rectangle is

(a=.40)times (a-.30) After simplification it gives a square -.02. The new area is 2% less than the original

Answer 2

Let the initial length be L.
And, the initial breadth be B.

Therefore, Initial Area = L × B

Now, Final length $= L+ \frac{40}{100} L$
                             $= 0.4 L$

And, Final breadth $= B - \frac{30}{100} B$ 
                              $= 0.7 B$

Thus, Final Area = 0.98 LB

As we can see, there is a decrease in the area.

So, Decrease in Area $= LB - 0.98 LB$

∴ % decrease in area $= \frac{0.02 LB}{LB}$ × $100$
                                   $= 2$% 

Or Simply:

Let the length of rectangle = 100 m
And, Let the breadth of rectangle = 10 m

Now, It's area = 1000 m²

Reduce the length by 30% to make it 70 m.
Increase the breadth by 40% to make it 14 m.

So, The altered area = 980 m²

Thus, The net reduction in Area $= \frac{1000 - 980}{1000}$ × 100
                                                   $= \frac{20}{1000}$ × 100                                                                                   $= 2$ %

∴ By reducing the length of the rectangle by 30% and increasing the breadth by 40%, the area reduces by 2%.

There's another small method:

Change in area is given by,
So, $A + B + \frac{AB}{100}$

⇒ $40 + (-30) + \frac{40 x -30}{100}$
⇒ $40 - 30 - 12$ 
  $= - 2$ %

Or, Area is reduced by 2%.

This method uses the direct relation, that's why it's a bit small. I recommend uh to not use in the exams as they will not entertain any method out of the textbook. The first 2 methods are perfect for exams though.

Hope This Helps :)