Question 806024
given:

 {{{P=42ft}}} and 

{{{A=108ft^2}}}

we know that {{{P=2L+2W}}} and {{{A=LW}}} where {{{L=length}}} and {{{W=width}}}

so, we have

{{{42ft=2L+2W}}}.........eq.1

{{{108ft^2=LW}}}..........eq.2
________________________________solve this system

{{{42ft=2L+2W}}}.........eq.1........solve for {{{L}}}

{{{42ft-2W=2L }}} 

{{{42ft/2-2W/2=L }}}

{{{21ft-W=L }}}.....eq.1a.......substitute in eq.2


{{{108ft^2=(21ft-W)W}}}..........eq.2

{{{108ft^2=21Wft-W^2}}}

{{{W^2-21Wft+108ft^2=0}}}......factor
 
{{{W^2-12Wft-9Wft+108ft^2=0}}}........group

{{{(W^2-12Wft)-(9Wft-108ft^2)=0}}}

{{{W(W-12ft)-9ft(W-12ft)=0}}}

{{{(W-12ft)(W-9ft) = 0}}}

solutions:

if 
{{{(W-12ft)  = 0}}} ==>{{{W=12ft}}}
if {{{ (W-9ft) = 0}}}==>{{{ W=9ft}}}

now find the length


{{{21ft-W=L }}}.....eq.1a ==>{{{W=12ft}}}


{{{21ft-12ft=L }}} 


{{{9ft=L }}} 

or

{{{21ft-W=L }}}.....eq.1a ==>{{{W=9ft}}}


{{{21ft-9ft=L }}} 


{{{12ft=L }}} 

so, the dimensions of a rug are: {{{L=12ft  }}} and {{{W=9ft}}} or vs