Chapter 9.1, Problem 55E

(a).

To determine

To Prove: The perimeter on an $n$ sided regular polygon inscribed inside a unit circle is $2n\sin \left(\frac{\pi }{n}\right)$ .

Explanation of Solution

Given Information: The regular polygon is $n$ sided and inscribed in a circle of radius $1$ .

Concept Used: The angle made by each side of an $n$ sided regular polygon at the center of the polygon is $\frac{2\pi }{n}$ .

In the figure shown below is an $n$ sided regular polygon inscribed in a unit circle .

  

In $\Delta OAB$ in the above figure,

 i $\overline{OA}=\overline{OB}=r=1$

 ii $OD$ is the perpendicular bisector from $O$ to $\overline{AB}$ .

 iii $\angle AOB=\alpha =\frac{2\pi }{n}$ and

 iv $\angle AOD$ is half of angle $\angle AOB$ .

So, $\angle AOD=\frac{\angle AOB}{2}=\frac{\alpha }{2}=\frac{\pi }{n}$

Therefore $AD=r\sin \angle AOD=1\cdot \sin \frac{\alpha }{2}=\sin \frac{\pi }{n}$

Since $\overline{OD}$ is the perpendicular bisector of $\overline{AB}$ , https://meet.google.com/ubp-hyod-ofn $AB=2\cdot AD=2\sin \frac{\pi }{n}$

Now, $\overline{AB}$ is one side of the $n$ sided regular polygon.

Therefore

  $Perimeter=a_n=n\cdot AB=n\times \left(2sin\frac{\pi }{n}\right)=2n\sin \frac{\pi }{n}$

(b).

To determine

To Calculate: The limit of a sequence.

Answer to Problem 55E

The sequence of perimeters of regular $n$ sided polygon converges to the limit $2\pi$ .

Explanation of Solution

Given Information: The limit to be calculated is $\underset{n\to \infty }{\lim }\left[2n\sin \left(\frac{\pi }{n}\right)\right]$

Concept Used:

• $\underset{\theta \to 0}{\lim }\frac{\sin \theta }{\theta }=1$
• If $\underset{x\to c}{\lim }f\left(x\right)=M$ , then $\underset{x\to c}{\lim }\left[k\cdot f\left(x\right)\right]=k\cdot M$( Constant Multiple Rule)

Calculation: In $\underset{n\to \infty }{\lim }\left[2n\sin \left(\frac{\pi }{n}\right)\right]$ , replace $\frac{\pi }{n}$ with $\theta$ to write

  $\underset{n\to \infty }{\lim }\left[2n\sin \left(\frac{\pi }{n}\right)\right]=\underset{\frac{\pi }{n}\to 0}{\lim }\left[\frac{2\pi \cdot \sin \left(\frac{\pi }{n}\right)}{\left(\frac{\pi }{n}\right)}\right]$

  $=\underset{\theta \to 0}{\lim }\left(\frac{2\pi \cdot \sin \theta }{\theta }\right)$

  $=2\pi \cdot \underset{\theta \to 0}{\lim }\left(\frac{\sin \theta }{\theta }\right)$ (Constant Multiple Rule)

  $=2\pi \cdot 1=2\pi$

Therefore, the sequence converges to the limit $2\pi$ , which is indeed the perimeter of the circle of unit radius in which the polygons were inscribed.