## Precalculus (6th Edition) Blitzer

Published by Pearson

# 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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