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Derivations of Pi and a Polygon of Degree n

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Date: 4 Mar 1995 18:05:14 -0500
From: MRS MAUREEN C HAMILTON
Subject: area of a polygon of degree n

I am curious about the derivation of pi.  I figured out it
was probably derived by computing the limit of the area of a
polygon of degree n as n approaches infinity.  For the life
of me I cannot remember the formula for deriving a polygon
of degree n.  Can you help?  This is my first E-mail message
on the Internet.
```

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Date: 5 Mar 1995 13:58:10 -0500
From: Dr. Ken
Subject: Re: area of a polygon of degree n

Hello there!

Wow, we're flattered that you chose US as the recipients of your first
e-mail message!

Yes, this is one method people have used to probe into the digits of Pi.
This method is one of the oldest methods (if not _the_ oldest) ways of
approximating Pi.  These days, though, people have figured out some
new ways that yield more digits faster, and these methods are pretty
complex.

Historically, most people have used the perimeter of a polygon instead
of its area to approximate Pi.  To do this, you would use the formula
Circumference = Pi * Diameter to approximate the digits of Pi.  To do
this, they inscribe a regular polygon in a circle of diameter 1.  Then
as the number of sides of the polygon gets big, its perimeter will
approach Pi.

If you want to use the formula Pi*r^2 to find an approximation to
Pi, you'd use a circle of radius 1, and inscribe a regular polygon
in it.  It's WAY helpful to draw a picture of what's going on.  You
can divide up the polygon into pie-wedges, and then you find the
area of each one and multiply by n.  So let's find the area of one of
the pieces.

We'll divide the triangle into 2 triangles by constructing the segment
from the center of the circle to the midpoint of the side of the polygon.
Then we've made a right triangle.  One of the 2 non-right angles
sticks out from the center of the circle; call this angle v.  Now the
lengths of the 2 legs of the triangle will be Sin[v] and Cos[v].  If you
don't see why, write us back (remember, the circle we've inscribe
this triangle in has radius 1).

So the area of the smaller triangle we made is Sin[t] * Cos[t] /2, so the
area of the larger triangle is Sin[t] * Cos[t].  Now, what is t in terms of
n?  It's 360/2n, i.e. 180/n.  Again, if you don't see why, ask us.  So the
area of this triangle is Sin[180/n] * Cos[180/n], and there are n such
triangles, so the total area of the polygon is n * Sin[180/n] * Cos[180/n].
If you take the limit of this term as n gets big, you should get Pi.

In fact, you could even simplify this term some.  If you just look at the
Cos[180/n] term, notice that 180/n is going to zero, and Cos[0] is 1.
So this won't contribute to the limit in the end, it'll just multiply it by 1.
So you could just use n*Sin[180/n] if you wanted to.

I hope this helps!

-Ken "Dr." Math
```
Associated Topics:
High School Geometry
High School Triangles and Other Polygons

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