Trapezium and Triangle

Since the trapezoid $ABCD$ is isosceles, let $AP = BP = x, CP = DP = y$ and angle $APB$ = angle $CPD$ = $\theta.$ The areas of triangles $APB, CPD$ are respectively given to be:

$\frac{1}{2} x^2 sin(\theta) = 4, \frac{1}{2} y^2 sin(\theta) = 9$

which solving for $y$ in terms of $x$ gives: $\frac{4}{x^2} = \frac{9}{y^2} \Rightarrow y = \frac{3}{2} x.$ The area of triangle $BPC$ now computes to:

$\frac{1}{2} xy sin(\pi - \theta) = \frac{1}{2} xy sin(\theta) = \frac{1}{2} x(\frac{3}{2} x) sin(\theta) = \frac{3}{2} \cdot [\frac{1}{2} x^2 sin(\theta)] = \frac{3}{2} \cdot 4 = \boxed{6}.;$

The triangles ABP and DCP are similar. We can pick base of DCP to be 6 (relative distribution of the area makes no difference) and get a height of 3.

To be similar, base of ABP has to be 4 and height 2.

The area of BCP can be calculated as half of height 5, obtained as 2+3, times the horizontal "base" going from P left to line BC.

The base is obtained by taking 2/5 of the difference DC - AB and adding it to AB. Then taking half of that. We obtain 2.4.

Area of $BCP =\frac12\times 2.4\times5=\boxed6$