Question 662382

given:

perimeter of a traffic sign is {{{P=100in}}}, 

its length {{{L}}} is {{{8in}}} longer than its width {{{W}}},=>..{{{L=W+8in}}}

...so, your traffic sign is a rectangle

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


and {{{100in=2(L+W)}}} ........plug in {{{L=W+8in}}} and solve for {{{W}}}


{{{100in=2(W+8in+W)}}}


{{{cross(100)50in=cross(2)(W+8in+W)}}}


{{{50in=2W+8in}}}


{{{50in-8in=2W}}}


{{{42in=2W}}}


{{{42in/2=W}}}


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


now find {{{L}}}


{{{L=W+8in}}}...plug in {{{W=21in}}}


{{{L=21in+8in}}}


{{{highlight(L=29in)}}}