The Math Forum

Ask Dr. Math - Questions and Answers from our Archives
Associated Topics || Dr. Math Home || Search Dr. Math

Areas of N-Sided Regular Polygon and Circle

Date: 02/24/2005 at 15:43:30
From: Rosanna
Subject: How the general formula for polygons turns into a circle.

In class we were discussing all the different possible polygon shapes
of land a farmer could have with 1000m of fencing, and which one would
give the largest area.  We discovered that as the number of sides
increased, so did the area.

My teacher mentioned that if you find the general formula for area of
a regular polgyon with a perimeter of 1000m, which is something like 
n(250000/Tan180), by manipulating it, you can turn it into the 
equation for the area of a circle.  I wondered if you could simply 
explain how this is possible.

I'm not sure how she got the general polygon formula, what the 250,000 
is.  Also she said you have to use radians instead of angles in the 
circle, and I am unsure how to do this.

Date: 02/24/2005 at 20:41:38
From: Doctor Rick
Subject: Re: How the general formula for polygons turns into a circle.

Hi, Rosanna.

I can't quite duplicate what your teacher said, but here are some 
related ideas.

If an N-gon (polygon with N sides) has perimeter P, then each of the 
N sides has length P/N.  If we connect two adjacent vertices to the 
center, the angle between these two lines is 360/N degrees, or 2*pi/N 
radians.  (Do you understand radian measure well enough to follow me 
this far?)

The two lines I just drew, plus the side of the polygon between them, 
form an isosceles triangle.  Adding the altitude of the isosceles 
triangle makes two right triangles, and we can use one of them to 
derive the equation

  sin(theta/2) =  s/(2R)

where theta is the apex angle (which I said is 2*pi/N radians), R is 
the length of the lines to the center (the radius of the circumscribed 
circle), and s is the length of the side (which I said is P/N).

Putting those values into the equation, we have

  sin(pi/N) = P/(2NR)

so that

  P = 2NR sin(pi/N)

gives the perimeter of the N-gon with circumradius R.

Can we see a connection between this formula and the perimeter of a 
circle?  The perimeter of the circumcircle is 2*pi*R.  As we increase 
N, the perimeter of the polygon should get closer and closer to this 
value.  Comparing the two, we see

  2NR*sin(pi/N) approaches 2*pi*R
  N*sin(pi/N)   approaches pi

You can check this out with a calculator, using big numbers for N such 
as your teacher's N=1000.  If you calculate the sine of an angle in 
degrees rather than radians, the formula will look like

  N*sin(180/N) --> pi

Now, let's back up and take another direction.  What is the AREA of 
the polygon?  I'll call the altitude of that triangle r (it is the 
radius of the inscribed circle).  Then the area of the triangle is 
half the altitude times the base, or rs/2.  The area of the polygon is 
N times the area of one triangle, since N triangles make up the 
polygon.  So the formula for the area of a regular polygon is simply

  Area = Nrs/2
       = rP/2

using the fact that P = Ns.  That's a neat formula: The area of a 
regular polygon is half the product of the perimeter and the inradius.

This relationship between perimeter and area is also true of a circle! 
The perimeter (circumference) of a circle is

  C = 2*pi*r

If I take half the product of the radius and the circumference, I get

  rC/2 = pi*r^2

which is the area of the circle.

Can I put these two lines of thought together?  I've got one problem 
remaining: One formula uses the inradius while the other uses the 
circumradius.  We can relate r and R by going back to that triangle: 
the ratio r/R = cos(theta/2) = cos(pi/N).  As N increases, this 
approaches 1, so that in the limit r=R; there is only one radius for 
a circle.  But because your formula involves the tangent, let's work 
with this.  Back to the perimeter formula:

  P = 2NR sin(pi/N)
    = 2N (r/cos(pi/N)) sin(pi/N)
    = 2Nr tan(pi/N)

so that

  r = P/(2N tan(pi/N))


  Area = rP/2
       = (P/(2N tan(pi/N)))P/2
       = P^2/(4N tan(pi/N))

This finally begins to look like what your teacher said.  If I put in 
P = 1000, I get

  Area = 250,000/(N tan(pi/1000)

If you calculate the tangent of an angle in degrees rather than 
radians, this will be

  Area = 250,000/(N tan(180/1000)

However, you can see that I said what I have to say about circles well 
before I got to this.  I don't know how your teacher was going to use 
this to talk about circles.  I hope this enlightens you a bit, anyway!

- Doctor Rick, The Math Forum 

Date: 02/25/2005 at 15:58:24
From: Rosanna
Subject: Thank you (How the general formula for polygons turns into a

Thank you very much, that really helped.  You do a great job!

- Rosanna
Associated Topics:
High School Conic Sections/Circles
High School Triangles and Other Polygons

Search the Dr. Math Library:

Find items containing (put spaces between keywords):
Click only once for faster results:

[ Choose "whole words" when searching for a word like age.]

all keywords, in any order at least one, that exact phrase
parts of words whole words

Submit your own question to Dr. Math

[Privacy Policy] [Terms of Use]

Math Forum Home || Math Library || Quick Reference || Math Forum Search

Ask Dr. MathTM
© 1994- The Math Forum at NCTM. All rights reserved.