Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 10 - Cumulative Review Exercises - Page 1128: 39

Answer

The width of the field is $50$ yards and the length is $100$ yards.

Work Step by Step

Let us assume the width of the field is $ x $. Then, the length would be $ x+50$. And equate the perimeter of the field as: $\begin{align} & 2\left[ x+\left( x+50 \right) \right]=300 \\ & 2\left[ 2x+50 \right]=300 \\ & 2x+50=150 \\ & 2x=150-50 \end{align}$ Then, simplify it further: $\begin{align} & x=\frac{100}{2} \\ & x=50 \end{align}$ Therefore, the width of the field is $50$ and the length is $50+50=100$. Thus, the width of the field is $50$ and the length is $100$.
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