Chapter 9.2, Problem 11Q

Solution

To find: the arc length and the area of a sector.

Arc length: $s=\frac{46\pi }{9}in$

Area: $A=\frac{184\pi }{9}i{n}^2$

Given information:

Given sector have radius 8 inches and the central angle $\theta =115°$ .

Definition Used:

Arc length and the area of a sector:

  

The arc length s and area A of a sector with radius r and central angle $\theta$ (measured in radians) are as follows.

Arc length: $s=r\theta$

Area: $A=\frac{1}{2}{r}^2\theta$

Property Used:

Degree can be converted into radian using:

  $1°=\frac{\pi }{180°}radian$

Explanation:

Here, the measure of central angle is given in degrees but as per the above definition this angle is required in radians. So, first step is to convert $\theta =115°$ to radians.

To convert $\theta =115°$ into radians, multiply it by $\frac{\pi }{180°}$ .

  $\begin{array}{c}115°=115°\times \frac{\pi }{180°}\\ =\frac{23\pi }{36}\text{ radian}\end{array}$

Now, use the formulas given in above definition to find the arc length and the area of the sector.

Here, use radius $r=8\text{ in}\text{.}$ and central angle $\theta =\frac{23\pi }{36}\text{ radian}$

Arc length:

  $\begin{array}{c}s=r\theta \\ =8\times \frac{23\pi }{36}\\ =\frac{46\pi }{9}\text{ in}\end{array}$

So, the arc length is $s=\frac{46\pi }{9}in$ .

Area:

  $\begin{array}{c}A=\frac{1}{2}{r}^2\theta \\ =\frac{1}{2}\times 8^2\times \frac{23\pi }{36}\\ =\frac{184\pi }{9}{\text{ in}}^2\end{array}$

So, the area is $A=\frac{184\pi }{9}i{n}^2$ .