Chapter 5.1, Problem 39E

(a)

To determine

To find: The area of the polygon for values of $\text{n=4}$ .

Answer to Problem 39E

Area of the polygon for values of $\text{n=4}$ is 2.

Explanation of Solution

Given information:

A regular $\text{n}$ side polygon inside a circle of radius $\text{1}$ .

Calculation:

The diagonal of the square is 2.

The side length of the square is $x=\sqrt{2}$

Since the area of the square is

  $\begin{array}{c}$ {\text{x}}^2\text{= }{(\sqrt{2})}^2$=2\end{array}$

Therefore the required area is 2.

(b)

To determine

To find: The area of the polygon for values of $\text{n=8}$ .

Answer to Problem 39E

Area of the polygon for values of $\text{n=8}$ is $2\sqrt{2}$

Explanation of Solution

Given information:

A regular $\text{n}$ side polygon inside a circle of radius $\text{1}$ .

Calculation:

Consider the octagon as a collection of 16 right triangles with a hypotenuse of length 1 and an acute angle measure $\frac{2\pi }{16}=\frac{\pi }{8}$

The height of the triangles are then $\cos \frac{\pi }{8}$ and the bases are $\sin \frac{\pi }{8}$

  

Area of the octagon $=16\times$ area of each triangle

  $\begin{array}{c}16\times \left(\frac{1}{2}\right)\times \left(\sin \frac{\pi }{8}\right)\times \left(\cos \frac{\pi }{8}\right)\\ =8\sin \frac{\pi }{8}\cos \frac{\pi }{8}\\ =4\sin \frac{\pi }{4}\text{ (Use the trig identity }\sin a\cos a=\frac{1}{2}\sin 2a\text{ ) }\\ =4\cdot \frac{\sqrt{2}}{2}\\ =2\sqrt{2}\end{array}$

The required area of the polygon for values of $\text{n=8}$ is $2\sqrt{2}$ .

(c)

To determine

To find: The area of the polygon for values of $n=16$

Answer to Problem 39E

Area of the polygon for values of $n=16$ is $3.061$

Explanation of Solution

Given information:

A regular $\text{n}$ side polygon inside a circle of radius $\text{1}$ .

Calculation:

Consider the octagon as a collection of 32 right triangles with a hypotenuse of length 1 and an acute angle measure $\frac{2\pi }{32}=\frac{\pi }{16}$

The height of the triangles are then $\cos \frac{\pi }{16}$ and the bases are $\sin \frac{\pi }{6}$

Area of the octagon $=32\times$ area of each triangle

  $\begin{array}{c}32\times \left(\frac{1}{2}\right)\times \left(\sin \frac{\pi }{16}\right)\times \left(\cos \frac{\pi }{16}\right)\\ =16\sin \frac{\pi }{16}\cos \frac{\pi }{16}\\ =8\sin \frac{\pi }{8}\text{ (Use the trig identity }\sin a\cos a=\frac{1}{2}\sin 2a\text{.) }\\ \approx 3.061\end{array}$

The required area of the polygon for values of $n=16$ is $3.061$ .

(d)

To determine

To compare: The areas in parts (a),(b) and (c) with the area of the circle.

Answer to Problem 39E

  $n$ increases, the area will approach $\pi$ .

Explanation of Solution

Given information:

A regular $\text{n}$ side polygon inside a circle of radius $\text{1}$ .

Explain:

The areas  found in (a), (b), and (c) are all smaller than $\pi$ , the area of the circle. However, when $n$ rises, the region will get closer to $\pi$ .