Algebra 2 (1st Edition)

Published by McDougal Littell
ISBN 10: 0618595414
ISBN 13: 978-0-61859-541-9

Chapter 9 Quadratic Relations and Conic Sections - Cumulative Review - Page 678: 41


The length is $115$ and the width is $75.$

Work Step by Step

Let $x$ be the length. Then the width is $x-40$. But we know that the circumference is $380$, thus: $x+(x-40)+x+(x-40)=380\\4x=460\\x=115$. Thus the length is $115$, and the width is $75.$
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