Chapter 6.1, Problem 40E

(a)

To determine

To calculate:The area of one of the n congruent triangles formed by drawing radii.

Answer to Problem 40E

The $\text{area of each isosceles triangle = }\frac{1}{2}\sin \frac{2\pi }{n}$

Explanation of Solution

Given information:

Inscribe a regular n-sided polygon inside a circle of radius 1 and compute the area of one of the n congruent triangles formed by drawing radii to the vertices of the polygon.

Formula used:

The isosceles triangles is made up of two right triangles having hypotenuse 1

Calculation:

Each of the isosceles triangles is made up of two right triangles having hypotenuse 1 and an acute angle measuring.

  $\frac{2\pi }{2n}=\frac{\pi }{n}$

  $\begin{array}{l}\text{length of perpendicular of right angled triangle}=\sin \frac{\pi }{n}\\ \text{length of base of right angled triangle}=\cos \frac{\pi }{n}\end{array}$

  $\begin{array}{l}\text{area of each isosceles triangle =}2\left(\frac{1}{2}\right)\left(\sin \frac{\pi }{n}\right)\left(\cos \frac{\pi }{n}\right)\\ \text{area of each isosceles triangle = }\frac{1}{2}\sin \frac{2\pi }{n}\\ \end{array}$

Conclusion:

The $\text{area of each isosceles triangle = }\frac{1}{2}\sin \frac{2\pi }{n}$

(b)

To determine

To calculate: The limit of an area of the inscribed polygon.

Answer to Problem 40E

The area of polygon is $\pi$

Explanation of Solution

Given information:

Inscribe a regular n-sided polygon inside a circle of radius 1

Formula used:

  $\text{area of polygon}=\frac{n}{2}\sin \frac{2\pi }{n}$

Calculation:

The area of polygon is n times the area of isosceles triangle.

  $\text{area of polygon}=\frac{n}{2}\sin \frac{2\pi }{n}$

Now n tends to infinity.

  $\begin{array}{l}\text{area of polygon}=\underset{n\to \infty }{\lim }\frac{n}{2}\sin \frac{2\pi }{n}\\ =\underset{n\to \infty }{\lim }\frac{\sin \frac{2\pi }{n}}{n}\pi \\ \text{area of polygon}=\pi \end{array}$

Conclusion:

The area of polygon is $\pi$

(c)

To determine

To calculate:The circle of area.

Answer to Problem 40E

The area of the polygon is $\pi r^2$

Explanation of Solution

Given information:

The n congruent triangles formed by drawing radii to the vertices of the polygon.

Formula used:

  $\text{area}\text{of}\text{polygon}=\frac{n}{2}{r}^2\sin \frac{2\pi }{n}$

Calculation:

In case of circle multiply each area by $r^2$

  $\begin{array}{l}\text{area of each isosceles traingle}=2r^2\left(\frac{1}{2}\right)\left(\sin \frac{\pi }{n}\right)\left(\cos \frac{\pi }{n}\right)\\ \text{area of each isosceles traingle}=\frac{1}{2}{r}^2\sin \frac{2\pi }{n}\end{array}$

  $\text{area}\text{of}\text{polygon}=\frac{n}{2}{r}^2\sin \frac{2\pi }{n}$

  $\begin{array}{l}\text{area}\text{of}\text{polygon}=\underset{n\to \infty }{\lim }\frac{n}{2}{r}^2\sin \frac{2\pi }{n}\\ =r^2\underset{n\to \infty }{\lim }\frac{\sin \frac{2\pi }{n}}{\frac{2\pi }{n}}\square \pi \\ \text{area}\text{of}\text{polygon}=\pi r^2\end{array}$

Conclusion:

The area of the polygon is $\pi r^2$