Find the area of the trapezium ABCD is trapezium AB||CD. 10 & 40 are the areas of the respective parts

How to find out the area of the trapezium?
 A: $$\triangle COD \sim \triangle AOB.$$
If $\dfrac{CD}{AB}=a$, then $\dfrac{\text{Area of}\space \triangle COD}{\text{Area of} \space \triangle AOB}=a^2$.
So, we conclude: $\dfrac{CD}{AB}=\sqrt{\frac{40}{10}}=\sqrt{4}=2$.
If we denote trapezium total height $h$, then 
heights of corresponding triangles are $\frac{2h}{3}$ and $\frac{h}{3}$ (all linear ratios of triangles $\triangle AOB$ and $\triangle COD$ are $\equiv 2$).
$\text{Area of} \space \triangle COD=\dfrac{CD}{2}\cdot \dfrac{2h}{3}=40$ $\implies$ $CD\cdot h=3\cdot 40=120$,
$\text{Area of} \space \triangle AOB=\dfrac{AB}{2} \cdot \dfrac{h}{3}=10$ $\implies$ $AB\cdot h = 6\cdot 10=60$;
Now Area of trapezium is
$$
\dfrac{AB+CD}{2}\cdot h = \frac{(120+60)}{2}=90.
$$
A: Let the diagonals intersect at P.
As pointed out, $\dfrac {DC}{AB} = \dfrac {2}{1}.$
Geometric theorems:-
I. If 2 ⊿s have the same altitude, then the ratio of their areas is equal to the ratio of their bases.
[Corollary of I]
II. If 2 ⊿s have the same altitude,their areas are the same when they have equal bases. 

By (II), area of ⊿PDA = area of ⊿PCB = x, say.
By (I), $\dfrac {40 + x}{10 + x} = \dfrac {2}{1}$
Then x = 20 and the result follows.
