Why is it that when comparing a squares area with a rectangles area, with the SAME PERIMETER, that the square has a greater area?

s = side length of square

4s = perimeter of square

L = length of rectangle

W = width of rectangle

P = 2L+2W = perimeter of rectangle

4s = 2L+2W .... since we want the perimeters the same

4s = 2(L+W)

s = 0.5(L+W)

s^2 = (0.5(L+W))^2 ... square both sides

s^2 = 0.25(L+W)^2

The last equation shows the area of the square, s^2, in terms of L and W. This way we can connect the square to the rectangle.

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The area of the rectangle is LW

Let's subtract this from the area of the square and see what we get

A = area of square = s^2

B = area of rectangle = LW

C = difference in area

C = A - B

C = s^2 - LW

C = 0.25(L+W)^2 - LW

C = 0.25(L^2+2LW+W^2) - LW

C = 0.25L^2+0.5LW+0.25W^2-LW

C = 0.25L^2-0.5LW+0.25W^2

C = 0.25(L^2-2LW+W^2)

C = 0.25(L-W)^2

Regardless of what we pick for L and W, the expression (L-W)^2 is always positive. Squaring a negative number leads to a positive result.

So C is always positive as long as

If C > 0, and C = A - B, then

C > 0

A-B > 0 .... replace C with A-B

A > B .... add B to both sides

This shows the square area is always larger than the rectangle.

Again this only works if L and W are different values. If L = W, then the argument falls apart because the rectangle becomes the square.

We make to have a nonsquare rectangle.

answer: for each 3 girls there is 2 boys . 3: 2

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