In the convex quadrilateral $A B C D$, the intersection point of the diagonals is $O$. What is the minimum area of the quadrilateral if the area of triangle $A O B$ is $4 \mathrm{~cm}^{2}$ and the area of triangle $C O D$ is $9 \mathrm{~cm}^{2} ?$
Let the areas of triangles $ABO, BCO, CDO, DAO$ be denoted by $T_{1}, T_{2}, T_{3}, T_{4}$ respectively.

![](https://cdn.mathpix.com/cropped/2024_05_02_5c64443451be33cf8af9g-1.jpg?height=499&width=859&top_left_y=225&top_left_x=622)

The first two triangles share the line segment $AO, OC$, so $T_{1}: T_{2}=AO: CO$. Similarly, for the last two triangles, $T_{3}: T_{4}=CO: AO$, thus $T_{1}: T_{2}=T_{4}: T_{3}$, or $T_{2} T_{4}=T_{1} T_{3}$.

In our case, $T_{1}=4 \mathrm{~cm}^{2}, T_{3}=9 \mathrm{~cm}^{2}$, so $T_{2} T_{4}=36 \mathrm{~cm}^{4}$. Since

\[
\left(T_{2}+T_{4}\right)^{2} \geq 4 T_{2} T_{4}
\]

$T_{2}+T_{4} \geq 2 \sqrt{T_{2} T_{4}}=12 \mathrm{~cm}^{2}$. Therefore, the area of the quadrilateral is at least $25 \mathrm{~cm}^{2}$. This area is exactly $25 \mathrm{~cm}^{2}$, for example, if $AC \perp BD$, and $AO=2 \mathrm{~cm}, BO=4 \mathrm{~cm}, CO=3 \mathrm{~cm}, DO=6 \mathrm{~cm}$, so a lower bound cannot be given.

Note. When a problem requires us to state a number that certain numbers cannot be smaller than, it is not always clear what should be considered a complete solution. Generally, in such problems, as in most similar problems, we should strive for the best result. However, it may not be possible to achieve the best result, or it may not be achievable with the available tools. In such cases, the solution that provides a higher lower bound is more valuable. Therefore, it is always worth examining whether the result we provide can be improved.