Start by stating the given info as mathematical expressions:
1. "The width of a rectangle is 6 cm less than its length." [tex]w=l-6[/tex]
2. "Its perimeter is 80 cm." [tex]P=80[/tex]
Using the formula for perimeter ([tex]P=2l+2w[/tex]) we can show: [tex]80=2l+2w[/tex].
We now have 2 equations in 2 unknowns and can solve by substitution: [tex]80=2l+2w[/tex] [tex]80=2l+2(l-6)[/tex] [tex]80=2l+2l-12)[/tex] [tex]92=4l[/tex] [tex]l=23[/tex]
The length is 23cm.
Having the length w can then find the width: [tex]w=l-6[/tex] [tex]w=23-6[/tex] [tex]w=17[/tex]
The width is 17cm.
And finally the area: [tex]A=lw[/tex] [tex]A=(23)(17)[/tex] [tex]A=391[/tex]
The area is [tex]391cm^{2}[/tex].
As a check we can use the known perimeter: [tex]80=2l+2w[/tex] [tex]80=2(23)+2(17)[/tex] [tex]80=46+34[/tex] [tex]80=80[/tex] Checks true.
