Question 1179263


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

area: {{{A=LW}}}

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

its area exceeds its perimeter by {{{14}}}

{{{LW=2(L+W)+14}}}..........substitute {{{L}}} from eq.1

{{{(W+3)W=2(W+3+W)+14}}}

{{{W^2+3W=2(2W+3)+14}}}

{{{W^2+3W=4W+6+14}}}

{{{W^2+3W-4W-20=0}}}

{{{W^2-W-20=0}}}

{{{(W - 5) (W + 4) = 0}}}

=>{{{W= 5}}} or {{{W= -4}}}-> disregard negative solution

go to

{{{L=W+3}}}........eq.1, substitute {{{W}}}

{{{L=5+3}}}

{{{L=8}}}

then perimeter is:

{{{P=2(L+W)}}}
{{{P=2(8+5)}}}
{{{highlight(P=26)}}} inches


check if area exceeds its perimeter by {{{14}}}: 

{{{A=LW}}}
{{{A=8*5}}}
{{{A=40}}}

{{{40-26=14}}}