5. The diagonals of an arbitrary trapezoid divide the trapezoid into four triangles. The areas of the triangles that contain the bases of the trapezoid are $P_{1}$ and $P_{2}$. Express the area of the trapezoid $P$ in terms of $P_{1}$ and $P_{2}$.
Solution. Let $O$ be the intersection of the diagonals, $P_{A B O}=P_{1}, P_{D O}=P_{2}, P_{A D O}=P_{4}$ and $P_{B C O}=P_{3}$. We will prove that the areas of triangles $A D O$ and $B C O$ are equal, i.e., that $P_{3}=P_{4}$. Indeed, triangles $A B D$ and $A B C$

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have a common base $A B$ and equal heights (the height of the trapezoid). Therefore, $P_{3}+P_{1}=P_{4}+P_{1}$, so $P_{3}=P_{4}$.

Triangles $A B O$ and $A D O$ have bases $B O$ and $D O$ and a common height $h^{\prime}$ from vertex $A$, so $P_{1}=\frac{1}{2} B O \cdot h^{\prime}$ and $P_{4}=\frac{1}{2} D O \cdot h^{\prime}$, hence $P_{1}: P_{3}=B O: D O$. Similarly, it can be shown that $P_{4}: P_{2}=B O: D O$ and since $P_{3}=P_{4}$, we get $P_{3}: P_{2}=B O: D O$. Therefore, $P_{1}: P_{3}=P_{3}: P_{2}$, from which it follows that $P_{3}=\sqrt{P_{1} P_{2}}$.

Since the area of the trapezoid is $P=P_{1}+P_{2}+2 P_{3}$, we have

$$
P=P_{1}+P_{2}+2 \sqrt{P_{1} P_{2}} \text {, i.e., } P=\left(\sqrt{P_{1}}+\sqrt{P_{2}}\right)^{2} \text {. }
$$