Chapter 6.2, Problem 32PFA

To determine

To find: Length of the rectangle .

Answer to Problem 32PFA

Length of rectangle = 26 inches.

Explanation of Solution

Given information:

Perimeter of rectangle = 72 inches.

Length = 6 + twice of width.

Perimeter of rectangle = $2\left(length+width\right)$

Let $l$ be the length of rectangle.

Let $w$ be the width of rectangle.

Therefore

  $\Rightarrow 2\left(l+w\right)=72$ -------(1)

Length = 6 + Twice of width

  $\Rightarrow l=6+2w$ ----------(2)

Substituting value of $l$ in equation-(1)

We get

  $\begin{array}{l}\Rightarrow 2\left(6+2w+w\right)=72\\ \Rightarrow 6+2w+w=36\\ \Rightarrow 6+3w=36\\ \Rightarrow 3w=36-6\\ \Rightarrow 3w=30\\ \Rightarrow w=\frac{30}{3}\\ \Rightarrow w=10\end{array}$

Substituting value of $w$ in equation-(2)

We get

  $\begin{array}{l}l=6+2w\\ \Rightarrow l=6+2\left(10\right)\\ \Rightarrow l=6+20\\ \Rightarrow l=26\end{array}$

Therefore

Length of rectangle = $l=26$ .

Width of rectangle = $w=10$ .