Derivations of Pi and a Polygon of Degree nDate: 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. 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 |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2013 The Math Forum
http://mathforum.org/dr.math/