Area of a right trapezium

In trapezium ABCD, $AB \parallel DC$ and $BC \perp DC$. Since $BC \perp DC$ and $AB \parallel DC$, $BC \perp AB$ as well. Height $h = BC = 4\text{ cm}$.

Draw perpendicular from D to AB meeting AB at point P. $DP = BC = 4\text{ cm}$ and $AP = DP / \tan 45^\circ = 4 / 1 = 4\text{ cm}$. Also $PB = DC = 5\text{ cm}$. So $AB = AP + PB = 4 + 5 = 9\text{ cm}$.

Area of trapezium = $\frac{1}{2} (AB + DC) \times h = \frac{1}{2} (9 + 5) \times 4 = 28\text{ sq. cm}$.