2. Let $O$ be the intersection point of the diagonals of trapezoid $A B C D$ $(A B \| C D)$, and $P_{1}$ and $P_{2}$ be the areas of triangles $A B O$ and $C D O$ respectively. Express the area of trapezoid $A B C D$ in terms of the areas $P_{1}$ and $P_{2}$.
Solution. Triangles $ABC$ and $ABD$ have equal areas because they have a common base and equal heights. Therefore, from

$$
P_{AOD} = P_{ABD} - P_{1} \quad P_{BCO} = P_{ABC} - P_{1}
$$

we get

$$
P_{ADO} = P_{BCO} = P_{3}
$$

![](https://cdn.mathpix.com/cropped/2024_06_05_b87d47f24deda06cfa28g-05.jpg?height=236&width=412&top_left_y=841&top_left_x=1087)

Since triangles $ADO$ and $ABO$ have equal heights, the ratio of their areas is equal to the ratio of their bases, i.e.,

$$
P_{3} : P_{1} = \overline{OD} : \overline{OB}
$$

Similarly, for the areas of triangles $CDO$ and $CBO$, we have

$$
P_{2} : P_{3} = \overline{OD} : \overline{OB}
$$

From (1) and (2), we get $P_{3} = \sqrt{P_{1} P_{2}}$, so for the area of the trapezoid, we have

$$
P = P_{1} + 2P_{3} + P_{2} = P_{1} + 2\sqrt{P_{1} P_{2}} + P_{2} = \left(\sqrt{P_{1}} + \sqrt{P_{2}}\right)^{2}
$$