Question 960068

let  the length be {{{L}}} and the width {{{W}}}
we know that a perimeter is 
{{{P=2(L+W)}}} 

if a rectangle has an perimeter of {{{56in}}}, then

{{{56in=2(L+W)}}}

{{{56in/2=L+W}}}

{{{28in=L+W}}}........solve for {{{W}}}

{{{W=28in-L}}}.....eq.1

if the width is one third of the length, then {{{W=(1/3)L}}}

substitute it in {{{W=28in-L}}}.....eq.1

{{{(1/3)L=28in-L}}}.......solve for {{{L}}}

{{{3(1/3)L=3*28in-3*L}}}

{{{L=84in-3L}}}

{{{L+3L=84in}}}

{{{4L=84in}}}

{{{L=84in/4}}}

{{{highlight(L=21in)}}}

now find {{{W}}}

{{{W=28in-L}}}

{{{W=28in-21in}}}

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

then, the area will be:

{{{A=L*W}}}

{{{A=21in*7in}}}

{{{highlight(A=147in^2)}}}