Chapter 5.1, Problem 28E

(a)

To determine

To show:

The expression of area of a polygon as $\frac{1}{2}n{r}^2\sin \left(\frac{2\pi }{n}\right)$.

Answer to Problem 28E

The area of a polygon $A_n=\frac{1}{2}n{r}^2\sin \left(\frac{2\pi }{n}\right)$ is proved.

Explanation of Solution

Given information:

The area of the polygon is $A_n$ (n-equal sides).

Polygon has n congruent triangles with central angle $\frac{2\pi }{n}$.

Calculation:

Procedure to draw one of the n congruent triangles in a circle is explained below:

• Let O be the center of the circle and AB be the one of the sides of the polygon.
• Draw the radius OC, so as to bisect the angle AOB. Also OC intersects AB at right angles and bisects AB.

Sketch one of the n congruent triangles in a circle with central angle $\frac{2\pi }{n}$ as shown in Figure 1.

Refer Figure 1.

The angle AOB is divided into 2 right angle triangles with legs of length $\frac{1}{2}\left(AB\right)=r\sin \left(\frac{\pi }{n}\right)\cdot r\cos \left(\frac{\pi }{n}\right)$

Therefore, the triangle $AOB$ has area of $2\frac{1}{2}\left[r\sin \left(\frac{\pi }{n}\right)\cdot r\cos \left(\frac{\pi }{n}\right)\right]$ (1)

Rearrange Equation (1) as shown below.

$\begin{array}{c}\text{Area}\text{of}\Delta AOB=\left[r\sin \left(\frac{\pi }{n}\right)\cdot r\cos \left(\frac{\pi }{n}\right)\right]\\ =r^2\sin \left(\frac{\pi }{n}\right)\cdot \cos \left(\frac{\pi }{n}\right)\\ =\frac{1}{2}{r}^2\sin \left(\frac{2\pi }{n}\right)\end{array}$

The area of the polygon with n equal sides $\left(A_n\right)$ is expressed as follows:

$A_n=n\times \text{Area}\text{of}\Delta AOB$ (2)

Substitute $\frac{1}{2}{r}^2\sin \left(\frac{2\pi }{n}\right)$ for $\text{Area}\text{of}\Delta AOB$ in Equation (2).

$\begin{array}{c}{A}_n=n\times \frac{1}{2}{r}^2\sin \left(\frac{2\pi }{n}\right)\\ =\frac{1}{2}n{r}^2\sin \left(\frac{2\pi }{n}\right)\end{array}$

Therefore, the area of a polygon with n equal sides $A_n$ is equal to $\frac{1}{2}n{r}^2\sin \left(\frac{2\pi }{n}\right)$.

Therefore, The expression of area of a polygon as $\frac{1}{2}n{r}^2\sin \left(\frac{2\pi }{n}\right)$.

(b)

To determine

To show:

The value of $\underset{n\to \infty }{\lim }{A}_n=\pi r^2$

Answer to Problem 28E

The value of $\underset{n\to \infty }{\lim }{A}_n=\pi r^2$ is proved.

Explanation of Solution

Refer part (a).

The area of a polygon with n equal sides is $A_n=\frac{1}{2}n{r}^2\sin \left(\frac{2\pi }{n}\right)$.

Find the value of $\underset{n\to \infty }{\lim }{A}_n$ as shown below:

Substitute $\left[\frac{1}{2}n{r}^2\sin \left(\frac{2\pi }{n}\right)\right]$ for $A_n$ in $\underset{n\to \infty }{\lim }{A}_n$.

$\begin{array}{c}\underset{n\to \infty }{\lim }{A}_n=\underset{n\to \infty }{\lim }\frac{1}{2}n{r}^2\sin \left(\frac{2\pi }{n}\right)\\ =\underset{n\to \infty }{\lim }\frac{1}{2}n{r}^2\frac{\sin \left(\frac{2\pi }{n}\right)}{\frac{2\pi }{n}}\frac{2\pi }{n}\\ =\underset{n\to \infty }{\lim }\frac{\sin \left(\frac{2\pi }{n}\right)}{\frac{2\pi }{n}}\pi r^2\end{array}$ (3)

Let $\theta =\frac{2\pi }{n}$.

As $n\to \infty$, $\theta \to 0$, Substitute $\frac{2\pi }{n}$ for $\theta$ in Equation (3).

$\begin{array}{c}\underset{n\to \infty }{\lim }\frac{\sin \left(\frac{2\pi }{n}\right)}{\frac{2\pi }{n}}\pi r^2=\underset{\theta \to 0}{\lim }\frac{\sin \left(\theta \right)}{\theta }\pi r^2\\ =\left(1\right)\pi r^2\\ =\pi r^2\end{array}$

Therefore, the value of $\underset{n\to \infty }{\lim }{A}_n$ is equal to $\pi r^2$.