Question 1141729

The perimeter of a rectangle is {{{22in}}}.

{{{P=2(L+W)}}}

=>{{{2(L+W)=22in}}}

=>{{{L+W=22in/2}}}

=>{{{L+W=11in}}}........eq.1


 If the length {{{L}}} is one more than the width {{{W}}}, we have

{{{L=W+1in}}}....eq.2


substitute it in eq.1

{{{W+1in+W=11in}}}........eq.1....solve for {{{W}}}

{{{2W=11in-1in}}}

{{{2W=10in}}}

{{{W= 5in}}}


go to eq.2, substitute {{{W}}}


{{{L=W+1in}}}....eq.2

{{{L=5in+1in}}}

{{{L=6in}}}


now find the area:


{{{A=L*W}}}

{{{A=6in*5in}}}

{{{A=30in^2}}}