Question

The perimeter of a rectangle is equal to 52.4 feet. The width is 11 less than twice the length. Find the length and width. “Both answer are decimal “

Answer 1

Step 1. Define the length

We will call the length of the rectangle x:

`\text{LENGTH}=x`

Step 2. Define the width

Since the width is 11 less than twice the length:

`\text{WIDTH}=2x-11`

Step 3. Use the formula for the perimeter of a rectangle:

`\text{PERIMETER}=2(WIDTH)+2(LENGTH)`

Substituting the value of the perimeter: 52.4, the width and length:

`52.4=2(2x-11)+2(x)`

Using distributive property on the right side:

`52.4=4x-22+2x`

And combining like terms in the right:

`52.4=6x-22`

Adding 22 to both sides:

`\begin{gathered} 52.4+22=6x-22+22 \\ 74.4=6x \end{gathered}`

Dividing both sides by 6:

`\begin{gathered} (74.4)/(6)=(6x)/(6) \\ 12.4=x \end{gathered}`

Thus, the width and length are:

`\begin{gathered} \text{LENGTH}=x=12.4ft \\ \text{WIDTH}=2x-11=2(12.4)-11=24.8-11=13.8ft \end{gathered}`