Question

The length of a rectangle is $\text{12 inches}$ more than its width. What is the width of the rectangle if the perimeter is $\text{42 inches}$ ?

Answer 1

Find the width of rectangle:

Let the width $w$ of the rectangle is $x\text{inches}$.

Then from the given condition length $l$ of the rectangle is $\left(x+12\right)\text{inches}$.

The perimeter $P$ of the rectangle is $42\text{inches}$.

It is known that the perimeter $P$ of the rectangle is :

$P=2\left(l+w\right)$

Then for the given rectangle :

$\begin{array}{rcl}2(x+12+x)& =& 42\\ & \Rightarrow & 2x+24+2x=42\\ \Rightarrow 4x& =& 42-24\\ \Rightarrow 4x& =& 18\\ \Rightarrow x& =& \frac{18}{4}\\ \Rightarrow x& =& 4.5\text{inches}\end{array}$

Therefor the width of the given rectangle is $\mathbf{4}\mathbf{.}\mathbf{5}\mathbf{}\mathbf{\text{inches}}$.