Question

Irrigation An irrigation system uses a straight sprinkler pipelong that pivots around a central point as shown. Because of an obstacle the pipe is allowed to pivot through${\mathbf{280}}^{\mathbf{°}}$ only. Find the area irrigated by this system.

Short answer

The area irrigated by this system is $\mathbf{219}\mathbf{,}\mathbf{911}{\mathbf{\text{ft}}}^{\mathbf{2}}$.

Step-by-step solution

Step 1. Given information.

The area A of a sector with central angle of radians is:

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

So use above formula to find the area irrigated by this system.
Given:
Central angle$\left(\theta \right)={280}^{°}$

radius$\left(r\right)=300ft$

To convert into radians, multiply the angle by$\frac{\pi }{180}$

Step 2. Finding the area irrigated by the system.

To convert $\left(\theta \right)$into radians, multiply by $\frac{\pi }{180}$:

$$\begin{gathered} \theta ={280}^{°}\times \frac{\pi }{{180}^{°}}\text{rad} \\ =\frac{14\pi }{9}\text{rad} \end{gathered}$$

The area irrigated by this system is:

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

Put given values in the above formula, we get:

$A=\frac{1}{2}\times {300}^2\times \frac{14\pi }{9}$

Putting $\pi =3.14$in the above formula, we get:

$$\begin{gathered} A=\frac{1}{2}\times {300}^2\times \frac{14\times 3.14}{9} \\ =\frac{14\times 3.14\times 300\times 300}{18} \\ =\frac{3956400}{18} \\ \approx 219911.485751 \\ \approx 219,911{\text{ft}}^2 \end{gathered}$$