[tex]Area: \ A = 350 \ m^2 \\ perimeter \ pathway: \ 90 \ m \\\\ we \ assume \ that \ the \ pathway \ is \ rectangular \\ width: \ w \\ length : \ l \\ \\Area: \ A= l\cdot w \\perimeter: \ P=2l + 2w[/tex]
[tex]350= l\cdot w \\90=2l + 2w \ \ /:2 \\\\350=l\cdot w \\45= l + w \\\\ l=45-w \\350=(45-w)\cdot w\\ \\350=45w-w^2\\w^2-45w+350=0[/tex]
[tex](w-10) (w-35)=0 \\w-10=0 \ \ or \ \ w-35 =0 \\w=10 \ m \ \ or \ \ w=35 \ m, \\\\w=10 \ m \ \ \ l=45-10=35 \ m \\or\\w=35, \ m \ \ \ l=45-35 =10 \ m\\\\Answer : \ Since \ this \ is \ a \ pathway, \ it \ makes \ sense \ that \ the \ width \ is \ the \ smaller \\ dimension, \ giving \ a \ width \ of \ 10 \ m \ and \ a \ length \ of \ 35 \ m[/tex]
