Question 727923


A rectangular parking lot has a length {{{L}}} that is {{{5m}}} more than twice its width {{{W}}}. 

so, we have that {{{L=2W+5m}}}...eq. 1

The area {{{A}}} of the parking lot is {{{63m^2}}}.


{{{A=L*W}}}...given {{{A=63m^2}}} and {{{L=2W+5m}}}, plug it in

{{{63m^2=(2W+5m)*W}}}....solve for {{{W}}}

{{{63m^2=2W^2+5mW}}}

{{2W^2+5mW-63m^2=0}}}.....use quadratic formula


{{{W = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}...note that {{{a=2}}}, {{{b=5m}}}, and {{{c=-63m^2}}}


{{{W = (-5m +- sqrt( (5m)^2-4*2*(-63m^2) ))/(2*2) }}}


{{{W = (-5m +- sqrt( 25m^2+504m^2 ))/4 }}}


{{{W = (-5m +- sqrt( 529m^2 ))/4 }}}


{{{W = (-5m +- 23m )/4 }}}....we need only positive root


{{{W = (-5m + 23m )/4 }}}


{{{W = 18m/4 }}}


{{{highlight(W =4.5m) }}}


now find the length


{{{L=2W+5m}}}...eq. 1


{{{L=2*4.5m+5m}}}


{{{L=9m+5m}}}


{{{highlight(L=14m)}}}