Introductory Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-805-X
ISBN 13: 978-0-13417-805-9

Chapter 2 - Section 2.5 - An Introduction to Problem Solving - Exercise Set - Page 166: 35

Answer

The length of the field is $360$ feet and the width is $160$.

Work Step by Step

The perimeter of a rectangle is given by the formula: $$P = 2l + 2w$$ We know from the problem that the perimeter is $1040$ feet, so we can plug that into the equation: $$1040 = 2l + 2w$$ We also know that the length is $200$ feet more than the width, $w$. We can plug this into the equation: $$1040 = 2(w + 200) + 2w$$ We simplify the equation: $$2w + 400 + 2w = 1040$$ We combine like terms: $$4w + 400 = 1040$$ We subtract $400$ from both sides: $$4w = 640$$ We divide both sides by $4$ to isolate $w$: $$w = 160$$ If we know that the width $w$ is $160$ feet, we know that the length $l$ is $200$ feet more than the width, so we have the length as: $$l = w + 200$$ We plug in $160$ for $w$: $$l = 160 + 200$$ We add the right-hand side of the equation: $$l = 360$$ The length of the field is $360$ feet and the width is $160$ feet.
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