Question 1048563
let the length be {{{L}}} and the width {{{W}}}

the area of rectangle is {{{A=L*W}}}
the perimeter of rectangle is {{{P=2L+2W}}}


if we have a rectangle with a area of {{{A=2888m^2}}} and {{{P=228m}}} for perimeter, than we have

 {{{L*W=2888m^2}}}......eq.1

{{{2L+2W=228m}}}......eq.2
------------------------------solve this system

start with {{{L*W=2888m^2}}}, solve for {{{L}}}

{{{L=2888m^2/W}}}.......substitute in eq.2


{{{2L+2W=228m}}}......eq.2..first simplify this one

{{{L+W=114}}}......substitute {{{L}}}

{{{2888/W+W=114}}}.......both sides multiply by {{{W}}}

{{{2888+W^2=114W}}}

{{{W^2-114W+2888=0}}}........factor

{{{W^2-38W-76W+2888=0}}}

{{{(W^2-38W)-(76W-2888)=0}}}

{{{W(W-38)-76(W-38)=0}}}

{{{(W-76) (W-38) = 0}}}

solutions:

{{{W=76m}}} or {{{W=38m}}}

now find {{{L}}}

{{{L=2888m^2/76m}}}->{{{L=38m}}}
{{{L=2888m^2/38m}}}->{{{L=76m}}}

so, 

{{{L=76m}}} and {{{W=38m}}} or the other way around