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;
[tex]l=2w-6[/tex]
Thus, the perimeter P of a rectangle is the sum of the outer boundaries of a shape. We have;
[tex]\begin{gathered} P=2w-6+w+2w-6+w \\ P=6w-12 \\ P=6(w-2) \end{gathered}[/tex]
Also, the area A of the rectangle is given as;
[tex]\begin{gathered} A=l\times w \\ A=(2w-6)w \\ A=2w^2-6w \\ A=2w(w-3) \end{gathered}[/tex]
