Chapter 9.1, Problem 29E

Solution

a.

To find: The half side x of the triangle in a hexagon and also the perimeter of hexagon.

Required value of x is $\frac{\text{1}}{\text{2}}\text{unit}$ and the perimeter of hexagon 6 units.

Given Data:

  

The hexagon is shown in Fig.1.

Method/Formula Used:

In a right angle triangle $\sin \theta =\frac{\text{height}}{\text{hypotenuse}}$ .

Calculations:

In right angle triangle OAM shown in Fig. 1,

  $\sin 30°=\frac{\text{AM}}{\text{OA}}$

Substitute x for AM, 1 for AM, and $1/2$ for $\sin 30°$ ,

  $\begin{array}{l}\frac{1}{2}=\frac{x}{\text{1}}\\ x=\frac{\text{1}}{\text{2}}\text{unit}\end{array}$

AB is the side of hexagon, also the chord of the circle of radius 1 in which the hexagon is inscribed.

  $\begin{array}{c}\text{AB=2×AM}\\ \text{=2×}\frac{1}{2}\\ =1\end{array}$

Thus, the side of the hexagon is 1, there are total 6 equal sides in the hexagon. Therefore, the perimeter p of the hexagon is,

  $\begin{array}{c}p\text{=6}\cdot \text{AB}\\ \text{=6}\cdot 1\\ =6\end{array}$

The perimeter of the hexagon is 6 units.

b.

To show: A regular n -sided polygon inscribed in a circle of radius 1 has perimeter of $2n\cdot \sin {\left(\frac{180}{n}\right)}^{°}$ .

Given data:

The polygon inscribed in a unit circle has n sides.

Method used:

The triangle ABC formed in an n -sided polygon inscribed in unit circle is makes angle $2\text{α}$ equals $360°/n$ shown in Fig.2.

  

Proof:

Let AB is one side of n-sided polygon. The side makes angle $2\text{α}$ at the center of the circle and polygon as well as shown in Fig.2. AB is the chord of the circle divided into two parts by normal AM. Therefore, in triangle OAM,

  $\begin{array}{c}\sin \text{α}=\frac{\text{AM}}{\text{OA}}\\ \text{AM}=\text{OA}\sin \text{α}\leftarrow \text{OA}=1\text{}\text{(radius}\text{}\text{of}\text{}\text{circle)}\\ \text{AM}=\text{1}\cdot \sin \text{α}\end{array}$

AB is the side of polygon given by

  $\begin{array}{c}\text{AB=2×sinα}\\ \text{=2sinα}\end{array}$

There are total n equal sides in the hexagon. Therefore, the perimeter p of the polygon is,

  $p=n\cdot AB$

Substitute $\text{2sinα}$ for AB,

  $p=n\cdot \text{2sinα}$

But $2\text{α}$ equals ${\left(\frac{360}{n}\right)}^{\text{ο}}$ , so substitute $\frac{1}{2}\cdot {\left(\frac{360}{n}\right)}^{\text{ο}}$ or ${\left(\frac{180}{n}\right)}^{\text{ο}}$ for $\text{α}$ ,

  $p=2n\cdot \text{sin}{\left(\frac{180}{n}\right)}^{\text{ο}}$

Hence, proved.

c.

To find: The result in part (b) in terms of $\text{π}$ and calculate results for n equals 50 or for 50-sided polygon.

Given:

The perimeter of n -sided polygon is $2n\cdot \text{sin}{\left(\frac{180}{n}\right)}^{\text{ο}}$ .

Method used:

The angle made by a circle at its center is ${\left(360\right)}^{\text{ο}}$ or $\text{2π}$ .

Calculations:

The perimeter p of n -sided polygon is

  $p=2n\cdot \text{sin}{\left(\frac{180}{n}\right)}^{\text{ο}}$ … (1)

Since, ${\left(360\right)}^{\text{ο}}$ equals $\text{2π}$ radians, therefore, ${\left(180\right)}^{\text{ο}}$ equals $\text{π}$ radians.

Substitute $\text{π}$ for ${\left(180\right)}^{\text{ο}}$ ,

  $p=2n\cdot \text{sin}\left(\frac{\pi }{n}\right)$ … (2)

When n is large, the angle $\frac{\pi }{n}$ is very-very small, so $\text{sin}\left(\frac{\pi }{n}\right)\approx \frac{\pi }{n}$ .

Substitute $\frac{\pi }{n}$ for $\text{sin}\left(\frac{\pi }{n}\right)$ in result (2),

  $\begin{array}{c}p=2n\left(\frac{\pi }{n}\right)\\ \text{=2π}\end{array}$

Thus, the perimeter of polygon is equal to the circumference of the unit circle.

The perimeter of the 50 sided polygon is,

  $\begin{array}{c}p=2\left(50\right)\cdot \text{sin}\left(\frac{\pi }{\text{50}}\right)\\ =6.28\text{units}\end{array}$