Chapter 6.1, Problem 61E

To determine

To find: the formula of area of segment..

Answer to Problem 61E

  $A=\frac{1}{2}{r}^2\left(\alpha -\text{sin}\alpha \right)$

Explanation of Solution

Given information:

The region bounded by an arc and its chord is termed as segment of circle.

  

  $\begin{array}{c}\text{Central angle}\left(\theta \right)=\alpha \text{ radian}\\ \text{Radius of circle}=r\end{array}$

Calculation:

Let area of triangle, sector and segment be represented as $A_t,A_s$ and $A$ respectively.

From the figure, it can be depicted that the area of sector is the sum of area of triangle and area of segment. This can be expressed as follows:

  $\begin{array}{c}{A}_s=A_t+A\\ or\\ A=A_s-A_t\left[\text{Rearranged}\right]\mathrm{...}\left(1\right)\end{array}$

Evaluate the area of sector as shown below.

  $\begin{array}{c}{A}_s=\frac{1}{2}{r}^2\theta \\ =\frac{1}{2}{r}^2\alpha \left[\text{Put the value of }\theta \right]\end{array}$

Evaluate the area of triangle as shown below.

  $\begin{array}{c}{A}_t=\frac{1}{2}\left(base\right)\left(height\right)\\ =\frac{1}{2}r\left(r\text{ sin}\alpha \right)\left[\text{Put the values}\right]\\ =\frac{1}{2}{r}^2\left(\text{sin}\alpha \right)\end{array}$

Put the values of $A_s$ and $A_t$ in equation (1) and evaluate the equation of area of segment as follows:

  $\begin{array}{c}A=\left(\frac{1}{2}{r}^2\alpha \right)-\left(\frac{1}{2}{r}^2\left(\text{sin}\alpha \right)\right)\\ =\frac{1}{2}{r}^2\left(\alpha -\text{sin}\alpha \right)\end{array}$

Thus the formula for calculating area of segment is $A=\frac{1}{2}{r}^2\left(\alpha -\text{sin}\alpha \right)$ .