Chapter 12.5, Problem 24WE

(a)

To determine

Find the perimeter of a regular polygon having n sides inscribed in a unit circle and perimeter of a regular polygon having n sides circumscribed about the unit circle.

Answer to Problem 24WE

  $2n\sin \left(\frac{180}{n}\right)$ , $2n\tan \left(\frac{180}{n}\right)$

Explanation of Solution

Formula Used:

Law of Sines:

  

  $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}$

Calculation:

The n -gon has n sides and if we join the sides to the centre of the circle, it forms n isosceles triangles with equal sides length 1 .

  

Since there are n equal triangles, the angle to the centre of each triangle is $\frac{360°}{n}$ .

Let base of each triangle is x .

In the isosceles triangle , the central angle is $\frac{360°}{n}$ , the other two angles are equal since the triangle is isosceles , so each angle is $90°-\frac{180°}{n}\mathrm{..............}[\text{sum of all angles is 180}°\text{]}$ .

Use the law of sines to find x :

  $\begin{array}{l}\frac{1}{\sin \left(90-\frac{180}{n}\right)}=\frac{x}{\sin \left(\frac{360}{n}\right)}\\ \Rightarrow x=\frac{\sin \left(2\cdot \frac{180}{n}\right)}{\cos \left(\frac{180}{n}\right)}\mathrm{.....................}[\text{multiply each side by }\sin \left(\frac{360}{n}\right)\text{ and sin(90-}\theta \text{)=cos}\theta \text{]}\\ \Rightarrow x=\frac{2\sin \left(\frac{180}{n}\right)\cos \left(\frac{180}{n}\right)}{\cos \left(\frac{180}{n}\right)}\mathrm{..................}[\sin 2\theta =2\sin \theta \cos \theta ]\\ \Rightarrow x=2\sin \left(\frac{180}{n}\right)\end{array}$

Since the n -gon has n sides , and one of the side is x , so , the perimeter of the n -gon inscribed inside a unit circle is nx :

  $nx=2n\sin \left(\frac{180}{n}\right)$

The n -gon circumscribed about the unit circle has n sides and if we join the sides to the centre of the circle, it forms n triangles:

  

Since there are n equal triangles, the angle to the centre of each triangle is $\frac{360°}{n}$ .

So, the central angle of the right angled triangle that bisects the triangle is $\frac{180°}{n}$

Let base of each right angled triangle is x and height is 1.

So,

  $\begin{array}{l}\tan \theta =\frac{side\text{ opposite to }\theta }{\text{side adjacent to }\theta }\\ \Rightarrow \tan \left(\frac{180}{n}\right)=\frac{x}{1}\\ \Rightarrow x=\tan \left(\frac{180}{n}\right)\end{array}$

Eace side of the n - gon is 2x = $2\tan \left(\frac{180}{n}\right)$

Since the n -gon has n sides , and one of the side is $2\tan \left(\frac{180}{n}\right)$ , so , the perimeter of the n -gon circumscribed about a unit circle is:

  $2n\tan \left(\frac{180}{n}\right)$

(b)

To determine

Find where the number is approached when n gets larger and larger:

  $n\sin \left(\frac{180}{n}\right)$ , $n\tan \left(\frac{180}{n}\right)$

Answer to Problem 24WE

both the numbers approaches 0 as n becomes larger and larger.

Explanation of Solution

Calculation :

As n becomes larger and larger , $\frac{180}{n}$ becomes smaller and smaller. So,

  $\begin{array}{l}n\to \infty \\ \Rightarrow \frac{180}{n}\to 0\\ \Rightarrow n\sin \frac{180}{n}\to 0\text{ and }n\tan \frac{180}{n}\to 0\end{array}$

Hence , both the numbers approaches 0 as n becomes larger and larger.