Area and Circumference of a Square

**citcat** · August 25th 2012, 03:51 PM

I know that the derivative of the area of a circle (pi*r^2) is the circumference of a circle (2pi*r). How can I show that relationship exists with the area and circumference of a square? (assuming the radius of a square is the distance from its centerpoint to each of its vertices)
Thanks!

**HallsofIvy** · August 25th 2012, 04:42 PM

If you know about circles, I would be very surprised if you did not know that the area of a square of side length s is $A= s^2$ and its 'circumference' (more commonly called 'perimeter' for everything except circles) is P= 4s. You also should know (using the Pythagorean theorem), that the length of a diagonal is $\sqrt{2}s$ and so what you are calling the 'radius' is half the diagonal, $r= \frac{\sqrt{2}}{2}s$ . From that, $s= r\sqrt{2}$ . Replacing s by that in the previous formulas, the area is $A= s^2= 2r^2$ and the perimeter is $4(r\sqrt{2})= 2(2\sqrt{2})r$ .

Notice that the constants multiplying $r^2$ in the area $2r$ in the perimeter are not the same. There is no single number like $\pi$ for squares.

**citcat** · August 25th 2012, 04:45 PM

So that relationship does not exist with squares?

**Plato** · August 25th 2012, 05:08 PM

Originally Posted by citcat

So that relationship does not exist with squares?

Well that depends upon what relationship means.

If $s$ is the length of a side then $C=4s~\&~A=s^2$ .

Then $\frac{A}{ds}=2s$ thus $2\frac{A}{ds}=C$ .

**Prove It** · August 25th 2012, 06:05 PM

Squares do not have a circumference.

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