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### Area and Perimeter in Polygons

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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

Sincerely,
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
13-year-old).

If you look at our FAQ on regular polygons

http://mathforum.org/dr.math/faq/formulas/faq.regpoly.html

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

Here's a sample polygon:

s
+-------+-------+
/ \      |      / \
/   \     |     /   \
/     \    |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

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

Substituting this into the area formula,

P^2
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
http://mathforum.org/dr.math/
```
Associated Topics:
High School Conic Sections/Circles
High School Geometry
High School Triangles and Other Polygons

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