Question 1193076

since given:

{{{AB = x/6+ 8}}}
{{{AD = x/3+ 5}}}
{{{BC = x - 2}}}
and
AB ≅ AD and BC ≅ DC , we have


{{{x/6+ 8=x/3+ 5}}}

{{{8-5=x/3-x/6  }}}

{{{3=2x/6-x/6  }}}

{{{3=x/6  }}}

{{{x=3*6}}}

{{{x=18}}}


then,


{{{AB = x/6+ 8=18/6+8=3+8=11}}}
{{{AD = x/3+ 5=18/3+5=6+5=11}}}

 {{{BC = x -2=18-2=16}}}
{{{BC = DC=16}}}


and, the perimeter of {{{ABCD}}} is:

{{{P=AB +AD+BC+DC}}}

{{{P=11 +11+16+16}}}

{{{P=54}}}