Find the area of trapezium $ABCD$ is a trapezium in which $AB||CD$. If $P$ is the point of intersection of diagonals $AC$ and $BD$ such that area of triangle $DPC=50cm^2$ and area of triangle $APB=32cm^2$.Then find area of trapezium $ABCD$.
I found that triangle APB and DPC are similar.The ratio of their sides is 4:5.The area of triangles DPC and CPB are equal.Please help me proceed. 

Consider the diagram below:

We are given that $\overline{AB} \parallel \overline{CD}$.  If two parallel lines are cut by a transversal, then alternate interior angles are congruent.  Thus, $\angle ABP \cong \angle CDP$ and $\angle CAB \cong \angle DCP$.  Therefore, $\triangle ABP \sim \triangle CDP$.  
We are also given that the area of $\triangle ABP$ is $32~\text{cm}^2$, while the area of $\triangle CDP$ is $50~\text{cm}$.  Hence, the ratio of the areas of the similar triangles is 
$$\frac{A(ABP)}{A(CDP)} = \frac{32~\text{cm}^2}{50~\text{cm}^2} = \frac{16}{25}$$
The ratio of the areas of similar triangles is the square of the ratio of corresponding sides.  Thus, 
$$\frac{|AB|}{|CD|} = \frac{|AP|}{|CP|} = \frac{|BP|}{|DP|} = \sqrt{\frac{16}{25}} = \frac{4}{5}$$
Let $E$ be the foot of the altitude from point $P$ to $\overline{AB}$; let $F$ be the foot of the altitude from point $P$ to $\overline{CD}$.  Since all right angles are congruent, $\angle BEP \cong \angle DFP$.  Since we have also established that $\angle EBP \cong \angle FDP$, $\triangle BEP \sim \triangle DFP$.  Thus, 
$$\frac{|EP|}{|FP|} = \frac{|BP|}{|DP|} = \frac{4}{5}$$
The area of trapezoid $ABCD$ is 
\begin{align*}
A(ABCD) & = \frac{1}{2}(|AB| + |CD|)(|EP| + |FP|)\\ 
        & = \frac{1}{2}\left(|AB| + \frac{5}{4}|AB|\right)\left(|EP| + \frac{5}{4}|EP|\right)\\
        & = \frac{1}{2}\left(\frac{9}{4}|AB|\right)\left(\frac{9}{4}|EP|\right)\\
        & = \frac{81}{16} \cdot \frac{1}{2} \cdot |AB| \cdot |EP|\\
        & = \frac{81}{16} \cdot A(\triangle ABP)\\
        & = \frac{81}{16}(32~\text{cm}^2)\\
        & = 162~\text{cm}^2
\end{align*}
