Precalculus (6th Edition) Blitzer

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

Chapter 8 - Section 8.5 - Determinants and Cramer's Rule - Exercise Set - Page 947: 77

Answer

The dimensions of the rectangles are 25 feet by 11 feet.

Work Step by Step

Consider the length and breadth of the rectangle to be $l$ and $b$ respectively. It is provided that the perimeter of the rectangle is $72$ feet. Consider the perimeter to be denoted by $p$, so: $p=2\left( l+b \right)$ Substitute the value of $p$ in the above equation: $\begin{align} & 72=2\left( l+b \right) \\ & l+b=36 \end{align}$ Also, it is provided that the length of the rectangle is 14 feet more than the width: $l=b+14$ So, $\begin{align} & l+b=36 \\ & b+14+b=36 \\ & 2b=36-14 \\ & b=11 \end{align}$ So, the length will be: $\begin{align} & l=b+14 \\ & =11+14 \\ & =25 \end{align}$ The dimensions of the rectangles are 25 feet by 11 feet.
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