I2.2 In the trapezium $A B C D, A B / / D C . A C$ and $B D$ intersect at $O$. The areas of triangles $A O B$ and $C O D$ are $P$ and 25 respectively. Given that the area of the trapezium is $Q$, find the value of $Q$.
$$
\begin{array}{l}
\frac{\text { area of } \triangle \mathrm{AOB}}{\text { area of } \triangle \mathrm{COD}}=\left(\frac{O A}{O C}\right)^{2} ; \frac{16}{25}=\left(\frac{O A}{O C}\right)^{2} \\
O A: O C=4: 5
\end{array}
$$
$\frac{\text { area of } \triangle \mathrm{AOB}}{\text { area of } \triangle \mathrm{BOC}}=\frac{4}{5}$ (the two triangles have the same height, but different bases.)
Area of $\triangle B O C=16 \times \frac{5}{4}=20$
Similarly, area of $\triangle A O D=20$
$Q=$ the area of the trapezium $=16+25+20+20=81$
$\triangle A O B \sim \triangle C O D$ (equiangular)