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Q82.

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Precalculus Mathematics for Calculus

Found in: Page 480

Book edition 7th Edition

Author(s) James Stewart, Lothar Redlin, Saleem Watson

Pages 948 pages

ISBN 9781337067508

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Short Answer

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 $280^{°}$ only. Find the area irrigated by this system.

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

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1. Given information.

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

$A = \frac{1}{2} r^{2} θ$

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

radius $(r) = 300 f t$

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

Step 2. Finding the area irrigated by the system.

To convert $(θ)$ into radians, multiply by $\frac{π}{180}$ :

$θ = 280^{°} \times \frac{π}{180^{°}} rad = \frac{14 π}{9} rad$

The area irrigated by this system is:

$A = \frac{1}{2} r^{2} θ$

Put given values in the above formula, we get:

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

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

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

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