Considering the triangle in the diagram
\(tan\theta = \frac{opp}{hyp}\\
Tan 63^{\circ} = \frac{h}{4}\\
h = 4 tan 63^{\circ}\\
Area of trapezium = \frac{1}{2}h(a+b)\\
\left(\frac{1}{2} \times 4 tan 63^{\circ}[5+9]\right)\\
=28\times 1.963 = 54.96 = 55cm^2\)