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Expert-verified Found in: Page 480 ### Precalculus Mathematics for Calculus

Book edition 7th Edition
Author(s) James Stewart, Lothar Redlin, Saleem Watson
Pages 948 pages
ISBN 9781337067508 # Irrigation An irrigation system uses a straight sprinkler pipe long 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. The area irrigated by this system is $\mathbf{219}\mathbf{,}\mathbf{911}{\mathbf{\text{ft}}}^{\mathbf{2}}$.

See the 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 $\frac{\pi }{180}$:

$\theta ={280}^{°}×\frac{\pi }{{180}^{°}}\text{rad}\phantom{\rule{0ex}{0ex}}=\frac{14\pi }{9}\text{rad}$

The area irrigated by this system is:

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

Put given values in the above formula, we get:

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

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

$A=\frac{1}{2}×{300}^{2}×\frac{14×3.14}{9}\phantom{\rule{0ex}{0ex}}=\frac{14×3.14×300×300}{18}\phantom{\rule{0ex}{0ex}}=\frac{3956400}{18}\phantom{\rule{0ex}{0ex}}\approx 219911.485751\phantom{\rule{0ex}{0ex}}\approx 219,911{\text{ft}}^{2}$ ### Want to see more solutions like these? 