# polygon

• Nov 28th 2007, 02:37 PM
harry1
polygon
A)Let An be the area of a polygon with n equal sides inscribed in a circle with radius r. By dividing the polygon into n congruent triangles with central angle 2pi/n, show that An=(1/2)nr^2sin(2pi/n).
B)Show that the limit as n approaches infinity = pir^2.
• Nov 28th 2007, 03:34 PM
galactus
The triangles forming the polygon are isosceles. The area of an isosceles triangle is given by $\frac{1}{2}r^{2}sin({\theta})$

The angle theta, in this case, is $\frac{2\pi}{n}$. And there are n of these triangles.

So, you have $\frac{1}{2}r^{2}\cdot{n}\cdot{sin(\frac{2\pi}{n})}$

Of course, as the number of triangles grows larger and larger, the area of the polygon will appraoach the area of the circle. This is similar to the method Archimedes used to find the area of a circle by inscribing a n-sided polygon.
Except, he didn't use trig. As far as we know.

$\lim_{n\rightarrow{\infty}}\frac{1}{2}r^{2}nsin(\f rac{2\pi}{n})$

$\frac{r^{2}}{2}\lim_{n\rightarrow{\infty}}nsin(\fr ac{2\pi}{n})$

Let's NOT use L'Hopital. No call for laziness.

Let $u=\frac{2\pi}{n}, \;\ n=\frac{2\pi}{u}$

$2{\pi}\lim_{u\rightarrow{0}}\frac{sin(u)}{u}$

Of course this famous limit is 1. And we have $2\pi$

Don't forget the $\frac{r^{2}}{2}$

Multiplying, $\frac{r^{2}}{2}\cdot{2\pi}={\pi}r^{2}$
• Nov 28th 2007, 04:03 PM
Soroban
Hello, Harry!

Quote:

A) Let $A_n$ be the area of a regular n-gon inscribed in a circle with radius $r.$

a) By dividing the polygon into $n$ congruent triangles with central angle $\frac{2\pi}{n}$,
show that: . $A_n\:=\:\frac{1}{2}\,nr^2\sin\left(\frac{2\pi}{n}\ right)$

b) Show that: . $\lim_{n\to\infty} A_n \;=\;\pi r^2$

Code:

              * * *           *          *         *              *       *                *       *        O        *       *        *        *       *    r  *  *  r    *             *  2π/n *       *  *          *  *       A* - - - - - - - *B           *          *               * * *

(a) One formula for the area of a triangle: . $A \:=\:\frac{1}{2}\,ab\sin C$
. . .(one-half the product of two sides times the sine of the included angle)

Hence, we have: . $A \;=\;\frac{1}{2}\,r^2\sin\left(\frac{2\pi}{n}\righ t)$

Therefore, the area of $n$ triangles is: . $\boxed{A_n \;=\;\frac{1}{2}\,nr^2\sin\left(\frac{2\pi}{n}\rig ht)}$

(b) Consider: . $\lim_{n\to\infty}A_n \;= \;\lim_{n\to\infty}\left[\frac{1}{2}\,nr^2\sin\left(\frac{2\pi}{n}\right)\r ight] \;=\;\frac{1}{2}r^2\cdot\lim_{n\to\infty}\left[n\sin\left(\frac{2\pi}{n}\right)\right]$ . [1]

Multiply by $\frac{2\pi}{2\pi}\!:\;\;\;\frac{2\pi}{2\pi}\cdot n\sin\left(\frac{2\pi}{n}\right) \;=\;2\pi\cdot\frac{n}{2\pi}\cdot\sin\left(\frac{2 \pi}{n}\right) \;=\;2\pi\cdot\frac{\sin\left(\dfrac{2\pi}{n}\righ t)}{\dfrac{2\pi}{n}}$

Let $\theta \,=\,\frac{2\pi}{n}$. . Note that, as $n\to\infty,\;\theta\to0$

Then [1] becomes: . $\frac{1}{2}\,r^2\cdot\lim_{\theta\to0}\left( 2\pi\cdot\frac{\sin\theta}{\theta}\right) \;=\;\pi r^2\cdot\underbrace{\lim_{\theta\to0}\frac{\sin\th eta}{\theta}}_{\text{This is 1}} \;=\;\pi r^2\cdot1 \;=\;\boxed{\:\pi r^2\:}$

Edit: Too fast for me, Galactus! . . . LOL!
.