If each side of a square is increased by x units then its perimeter is increased by how many units?

Expert Answers
hala718 eNotes educator| Certified Educator

Let the length of the square's side be A.

We know that the perimeter of the square equals the sum of all sides = 4* side.

==> Then, let the perimeter be P1.

P1 =  4A....................(1)

Now if the side length increased by X:

Then, the length sides = (x+A)

Then the perimeter (P2) is given by: 

P2 = 4(x+ A).

Open brackets.

   = 4x + 4A

But we know that P1 = 4A.

==> P2 = 4x + P1

Then, the perimeter will increase by 4x.

giorgiana1976 | Student

Let's note the length of the side of the square as l.

As we know, the lengths of the sides of the square are equal.

The perimeter of the square is the sum of the lengths of the sides. Since the lengths are equal, we'll get:

P1 = 4*l

If each side is increased by x units, we'll have the new perimeter;

P2 = (l+x) + (l+x) + (l+x) + (l+x)

P2 = 4(l+x)

Now, the increasing of P2 over P1 is:

P2 - P1 = 4(l+x) - 4l

We'll remove the brackets and we'll get:

P2 - P1 = 4l + 4x - 4l

We'll eliminate like terms:

P2 - P1 = 4x

The original perimeter is 4x units smaller than the new one.

neela | Student

Since there are 4 equal sides for a square, the perimeter of the square is 4 times the side. So if the side of a square measures a units, then the perimeter p of the square is given by:

p = 4a.

 Now each side is increased by x units. So the each side is a+x units now. Then the new perimeter p' increase in the

p' = 4(a+x).

Therefore increse in perimeter = p'-p = 4(a+x)-4a = 4a+4x-4a = 4x.

Therefore the increase in the perimeter  = 4x.