Question 866446
let the length of a rectangle be {{{L}}}

let the width of a rectangle be {{{W}}}

by definition: 
the perimeter is {{{P=2L+2W}}} and the area is {{{A=L*W}}}

given: the length of a rectangle {{{L}}} is {{{3ft}}} longer then its width {{{W}}} => {{{L=W+3ft}}}......eq.1

if the perimeter of the triangle is {{{P=34ft}}} then we have: 

{{{P=2L+2W}}} ...substitute {{{L}}} and {{{P}}}

{{{34ft=2(W+3ft)+2W}}} ...solve for {{{W}}}

{{{34ft=2W+6ft+2W}}}

{{{34ft-6ft=4W}}}

{{{28ft=4W}}}

{{{28ft/4=W}}}

{{{highlight(7ft=W)}}}

now find {{{L}}}


{{{L=W+3ft}}}......eq.1

{{{L=7ft+3ft}}}

{{{highlight(L=10ft)}}}

finally, you can find the area:

 {{{A=L*W}}} ...............plug in {{{highlight(L=10ft)}}} and {{{highlight(7ft=W)}}}


{{{A=10ft*7ft}}}

{{{highlight(A=70ft^2)}}}