ABCD is a trapezium with $\overline{AB}$  parallel to $\overline{CD}$  and the diagonals intersect at the point P. The area of $\Delta ABP$  is 32 $c{m}^2$  and the area of $\Delta ADP$  is 50 $c{m}^2$ . The area of the trapezium is (in sq. cm).

$\frac{DP}{PB}=\frac{y}{32}=\frac{50}{x}$

$\frac{AP}{AC}=\frac{32}{x}=\frac{y}{50}$

We know that, In trapezium,

Diagonals interest proportionally.

$\therefore \frac{DP}{PB}=\frac{CP}{AP}$

$\therefore \frac{50}{x}=\frac{x}{32}$ and $\frac{y}{32}=\frac{50}{y}$

$x^2=50\times 32$

$x=40$

$y^2=32\times 50$

$y=40$

Area of trapezium $=50+32+x+y$

$=50+32+40+40$

$=162$ sq.cm