Area of Trapezium

Since $\angle BAD = 90^\circ$ and $AD \parallel BC$, $AB$ is perpendicular to both $AD$ and $BC$. Let $AB = h$ (height of trapezium) and $CD = c$.

Perimeter $= AB + BC + CD + AD = h + 3 + c + 8 = h + c + 11 = 36$. So, $h + c = 25 \implies c = 25 - h$.

Drop a perpendicular from $C$ to $AD$ meeting $AD$ at $E$. Then $AE = BC = 3\text{ cm}$, so $ED = AD - AE = 8 - 3 = 5\text{ cm}$. Also $CE = AB = h$. In right triangle $CED$: $CE^2 + ED^2 = CD^2$ $h^2 + 5^2 = c^2$ $h^2 + 25 = (25 - h)^2$ $h^2 + 25 = 625 - 50h + h^2$ $50h = 600 \implies h = 12\text{ cm}$

Area of trapezium $ABCD = \frac{1}{2} \times (AD + BC) \times h = \frac{1}{2} \times (8 + 3) \times 12 = 66\text{ sq.cm}$.