This can be illustrated with formula and with a an example.

Let side of a square be x.

Let the length and breadth of a rectangle be l and b whose area is equal to the area of the square.

I am taking few value of side of the square and taking the equal area the length and breadth are set a value.

Case 1:

let x=2

Therefore X^{2} = 4

Let l=4

and b=1

So that the area remain same.

suppose the perimeter of the rectangle be greater than the perimeter of the square.

Therefore

2(l+b)-4x will be positive.

2(4+1)-4*2

=10-8

=2

Case 2:

let x=4

Therefore X^{2} = 16

Let l=8

and b=2

So that the area remain same ie 16 unit.

Let check again:

2(l+b)-4x will be positive.

2(8+2)-4*4

=20-16

=4

Case 3:

let x=8

Therefore X^{2} = 64

Let l=16

and b=4

So that the area remain same ie 64 unit.

Let check again:

2(l+b)-4x will be positive.

2(16+4)-4*8

=40-32

=8

Therefore in all three cases we noticed that the perimeter of the rectangle is always positive.

Therefore if the area of a square and that of a rectangle are equal then the perimeter of the rectangle will always be greater.

Please comment if you are satisfied with the answer.

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