$\text{The area of a trapezium is given by:}$
$A=\frac{1}{2}\times \text{(Sum of parallel sides)}\times \text{(height)}$
$\Rightarrow 36=\frac{1}{2}\times (12+6)\times h$
$\Rightarrow h=4cm$
$\text{So now in the trapezium,}$​
$h=4cm$
$a=b=6cm$
$AB=CD\text{[S-A-S congruency in}△AFB\text{and}△DEC]$
$\therefore 2AB+a=12cm$
$\Rightarrow 2AB=12-6$
$\Rightarrow AB=CD=3cm$
$\text{So now in}△ABF,A{B}^2+h^2=d^2$​
$d^2=3^2+4^2=25$​
$d=5cm$
As the trapezium is isosceles, the slant sides of the trapezium are equal in length
$d=c=5cm$
$\therefore$ the perimeter of the trapezium
$=(2\times AB)+a+c+b+d$
$=(2\times 3)+6+5+6+5=6+22=28cm$