Precalculus (6th Edition) Blitzer

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

Chapter 7 - Cumulative Review Exercises - Page 880: 34

Answer

The dimensions are 4 m by 9 m.

Work Step by Step

Let us consider the length as $ L $ and width as $ W $. Therefore, from the given conditions, $\begin{align} & LW=36\ \text{square meters} \\ & L=2W+1 \end{align}$ Therefore, put the value of $ W $ from equation $ L=2W+1$ in equation, $ LW=36$. So, $\begin{align} & LW=36 \\ & \left( 2W+1 \right)W=36 \\ & 2{{W}^{2}}+W-36=0 \end{align}$ Solving further $\begin{align} & W(2W+9)-4(2W+9)=0 \\ & (W-4)(2W+9)=0 \end{align}$ And, $ W=4,W=\frac{-9}{2}$ Take the positive value of the width. Now, find the value of the length. $\begin{align} & W=2W+1 \\ & =2\times 4+1 \\ & =9\ \text{m} \end{align}$ Thus, the dimension of the rectangle is 4 meters by 9 meters.
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