It is given in the question that the length of the rectangle is 6 less than double the width. Thus, we have the diagram as;
Where the length l is;
![l=2w-6](https://img.qammunity.org/2023/formulas/mathematics/college/a5pa64dtlsf98k4vgqxz4odav1ddun0luj.png)
Thus, the perimeter P of a rectangle is the sum of the outer boundaries of a shape. We have;
![\begin{gathered} P=2w-6+w+2w-6+w \\ P=6w-12 \\ P=6(w-2) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/g1lmxocpg6qtnb93szvq9u4tv6fxog0ayw.png)
Also, the area A of the rectangle is given as;
![\begin{gathered} A=l* w \\ A=(2w-6)w \\ A=2w^2-6w \\ A=2w(w-3) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/j7atev81uayg8qufrehptj8wbh4zjw27k9.png)