Why is the area of the square larger than the area of the rectangle when perimeter is same?

Question

I know that this is true, but how do I explain it?  You have a square and a rectangle with the same perimter, but the area of a square is larger.  Why?

Answer 1

Suppose the side of the square is a then its area is a². Its perimeter is 4a.

Now consider a rectangle with length L and width W, perimeter=2(L+W)=4a because it has the same perimeter as the square.

So L+W=2a, and W=2a-L. Now suppose L=a+d. W=2a-(a+d)=2a-a-d=a-d. Now let's just check the perimeter:

2(L+W)=2(a+d+a-d)=4a. So that checks out. What about the area? The area of the rectangle is (a+d)(a-d)=a²-d².

This value is always less than a², so, unless d=0, the area of the rectangle is always less than a square of the same perimeter.