## Elementary Technical Mathematics

Published by Brooks Cole

# Chapter 16 - Cumulative Review - Page 568: 9

#### Answer

The length of the rectangle is $15\text{m}$ and the width is $4.9\text{m}$.

#### Work Step by Step

Suppose the width of the rectangle is $b=x$, then the length of the rectangle is $l=x+10.1$. The perimeter of the rectangle is $P=2\left( l+b \right)$; therefore, \begin{align} & P=2\left( l+b \right) \\ & 39.8=2\left( x+10.1+x \right) \\ & 39.8=2\left( 2x+10.1 \right) \\ & 39.8=4x+20.2 \end{align} Further simplify: \begin{align} & 39.8-20.2=4x \\ & 4x=19.6 \\ & x=\frac{19.6}{4} \\ & x=4.9 \end{align} Therefore, the width of the rectangle is $x=4.9$ and the length will be, \begin{align} & l=x+10.1 \\ & =4.9+10.1 \\ & =15.0 \end{align} Hence, the length of the rectangle is $\text{15m}$ and the width is $\text{4}\text{.9m}$.

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