# Area and Circumference of a Square

• Aug 25th 2012, 03:51 PM
citcat
Area and Circumference of a Square
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!
• Aug 25th 2012, 04:42 PM
HallsofIvy
Re: Area and Circumference of a Square
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 $\displaystyle A= s^2$ 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 $\displaystyle \sqrt{2}s$ and so what you are calling the 'radius' is half the diagonal, $\displaystyle r= \frac{\sqrt{2}}{2}s$. From that, $\displaystyle s= r\sqrt{2}$. Replacing s by that in the previous formulas, the area is $\displaystyle A= s^2= 2r^2$ and the perimeter is $\displaystyle 4(r\sqrt{2})= 2(2\sqrt{2})r$.

Notice that the constants multiplying $\displaystyle r^2$ in the area $\displaystyle 2r$ in the perimeter are not the same. There is no single number like $\displaystyle \pi$ for squares.
• Aug 25th 2012, 04:45 PM
citcat
Re: Area and Circumference of a Square
So that relationship does not exist with squares?
• Aug 25th 2012, 05:08 PM
Plato
Re: Area and Circumference of a Square
Quote:

Originally Posted by citcat
So that relationship does not exist with squares?

Well that depends upon what relationship means.

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

Then $\displaystyle \frac{A}{ds}=2s$ thus $\displaystyle 2\frac{A}{ds}=C$.
• Aug 25th 2012, 06:05 PM
Prove It
Re: Area and Circumference of a Square
Squares do not have a circumference.