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Area and Perimeter in Polygons

Date: 06/24/99 at 15:15:39
From: Mike Hansen
Subject: Geometry

Dear Dr. Math,

I teach Algebra I to some very precocious 13-year-olds, and one in 
particular who came up with a formula for area that I would like you 
to look over for me.

According to James, he saw a pattern in the areas of regular polygons 
and decided to dedicate some time to trying to find a formula that 
would calculate the areas of regular polygons when the lengths of the 
sides and the number of sides is known. He sent me the following:

This formula finds the area of any regular polygon where a is the 
length of each side, and b is the number of sides.

     A = [((1/2)a) / (tan(360/2b))] * (1/2)a * b

With James' permission I rewrote the formula as the following:

     A = (a^2b) / ((4tan(360/2b))

The formula works perfectly. Because of my general belief that seldom 
will any of us have an original thought I would like to pose the 
following question to you:

Do you recognize either form of this formula? Have you seen a general 
formula for areas of regular polygons that is similar? Is there any 
historic precedent for this formula?

I appreciate any time you devote to this query.  I respectfully await 
your response.

Mike Hansen

Date: 06/24/99 at 23:37:25
From: Doctor Peterson
Subject: Re: Geometry

Hi, Mike.

Yes, this is a good formula, and not too unusual (except coming from a 

If you look at our FAQ on regular polygons   

you will find a formula

     K = na^2 cot(alpha/2)/4

where n is the number of sides and a is the length of a side. The 
angle alpha is 360/n. If we rewrite cot as 1/tan, this becomes

             n a^2
     K = ------------
         4 tan(180/n)

which is exactly the formula you give. (I wrote 180/n rather than 
360/2n to avoid ambiguity and the need to write 360/(2n) for clarity.)

You may be interested in the following answer I gave a few days ago, 
in the course of which I derived essentially this formula, except that 
I used the perimeter P = na rather than a. James probably did it more 
or less the same way. The result is pretty neat; if you graph it you 
see that it approaches 1 as N increases.

James sounds like me at that age, and my son too, who is now 14. 
Encourage him!

>Dear Dr. Math,
>A circle is the largest area you can have with a given circumference. 
>Any regular polygon takes up X% of the area of a circle with the same 
>circumference/perimeter. I really need a formula where I could plug 
>in the number of sides of a regular polygon and see what % of a 
>circle it covers if they have the same perimeter.  Please help me.

Here's a sample polygon:

             / \      |      / \
            /   \     |     /   \
           /     \    |a   /     \
          /       \   |   /       \
         /         \  |  /         \
        /           \ | /           \
       /             \|/2pi/N        \
       \             / \             /
        \           /   \           /
         \         /     \         /
          \       /       \       /
           \     /         \     /
            \   /           \   /
             \ /             \ /

The perimeter of an N-gon with apothem a is N times the side length s:

     P = N * s = N * 2 * a * tan(pi/N)

The area will be N times the area of one triangle:

     A = N * 1/2 * s * a = N * a * a * tan(pi/N)

Solving the first for a in terms of P, we get

     a = ------------
         2N tan(pi/N)

Substituting this into the area formula,

     A_p = -------------
           4 N tan(pi/N)

A circle with the same perimeter will have radius r = P/(2 pi), so the 
area is

                    pi P^2   P^2
     A_c = pi r^2 = ------ = ----
                    4 pi^2   4 pi

Divide these and you're done:

                      P^2       4 pi       pi
     A_p/A_c =  ------------- * ---- = -----------
                4 N tan(pi/N)   P^2    N tan(pi/N)

- Doctor Peterson, The Math Forum   
Associated Topics:
High School Conic Sections/Circles
High School Geometry
High School Triangles and Other Polygons

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