Question 672459

a perimeter of the rectangle: {{{P=2L+2W}}}

if the length {{{L=W+80mil}}}

and {{{P=1200mil}}}

then

{{{1200mil=2(W+80mil)+2W}}}......solve for {{{W}}}

{{{1200mil=2W+160mil+2W}}}

{{{1200mil-160mil=4W}}}

{{{1040mil=4W}}}

{{{1040mil/4=W}}}

{{{highlight(260mil=W)}}}

now find {{{L}}}


{{{L=W+80mil}}}

{{{L=260mil+80mil}}}

{{{highlight(L=340mil)}}}