Question

The length of a rectangle is six feet less than twice the width. If the perimeter of the rectangle is 48, find both dimensions

Answer 1

Width=10 ft

Lenth=14 ft

Explanation

Step 1

Let

length(L)

Width(W)

The length of a rectangle is six feet less than twice the width,it is

$\begin{gathered} L=2W\text{ -6 Equation(1)} \\ \end{gathered}$

Also, the perimeter of the rectangle is 48,the perimeter is given by:

$\begin{gathered} \text{Perimeter}=2L+2W \\ 48=2L+2W\text{ Equation(2)} \end{gathered}$

Step 2

using equation (1) and (2), find L and W

a)

replace equation (1) in equation (2)

$\begin{gathered} 48=2L+2W \\ 48=2(2W-6)+2W \\ 48=4W-12+2W \\ 48=6W-12 \\ 6W=60 \\ W=10\text{ ft} \end{gathered}$

b) replace the value fo W in equation (1) to find L

$\begin{gathered} L=2W\text{ -6} \\ L=2\cdot10\text{ -6} \\ L=20-6 \\ L=14 \end{gathered}$

I hope this helps you