6. In triangle $ABC$, lines parallel to the sides of the triangle are drawn through an arbitrary point $O$. As a result, triangle $ABC$ is divided into three parallelograms and three triangles. The areas of the resulting triangles are $6 \text{ cm}^2$, $24 \text{ cm}^2$, and $54 \text{ cm}^2$. Find the area of triangle $ABC$.

$$
\text { (20 points) }
$$
# Solution:

The triangles are similar. Let the first and third be similar with a similarity coefficient $k_{1}$, and the second and third with a similarity coefficient $k_{2}$. Then for the areas of these triangles, we have the ratios $\frac{S_{1}}{S_{3}}=k_{1}^{2}, \frac{S_{2}}{S_{3}}=k_{2}^{2}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_009f2643c33564a3e4a2g-3.jpg?height=391&width=757&top_left_y=1895&top_left_x=1249)

The areas of the parallelograms are expressed as follows: $S_{1 p}=k_{1} x h=2 k_{1} S_{3}, S_{2 p}=k_{2} x h=2 k_{2} S_{3}, S_{3 p}=k_{1} k_{2} y z \sin \alpha=2 k_{1} k_{2} S_{3}$. $S_{A B C}=S_{1}+S_{2}+S_{3}+2\left(k_{1}+k_{2}+k_{1} k_{2}\right) S_{3}=S_{1}+S_{2}+S_{3}+2\left(\sqrt{\frac{S_{1}}{S_{3}}}+\sqrt{\frac{S_{2}}{S_{3}}}+\frac{\sqrt{S_{1} S_{2}}}{S_{3}}\right) S_{3}=$ $=S_{1}+S_{2}+S_{3}+2 \sqrt{S_{1} S_{3}}+2 \sqrt{S_{2} S_{3}}+2 \sqrt{S_{1} S_{2}}=6+24+54+2 \sqrt{6 \cdot 54}+2 \sqrt{24 \cdot 54}+2 \sqrt{6 \cdot 24}=216$.

Answer: 216.

## Solution for Variant #2