7. The area of quadrilateral $\mathrm{ABCD}$ is $48 \mathrm{~cm}^{2}$ and point $\mathrm{D}$ lies on segment $\overline{\mathrm{BE}}$ such that $|E D|:|D B|=1: 2$. What is the area of triangle $\triangle \mathrm{ABC}$?

![](https://cdn.mathpix.com/cropped/2024_06_03_a89fbb865052a7f27847g-08.jpg?height=485&width=580&top_left_y=791&top_left_x=561)
7. Since $|D B|=2|E D|$, then $P_{\triangle D B C}=2 \cdot P_{\triangle E D C}$ and $P_{\triangle A B D}=2 \cdot P_{\triangle A D E}$. 4 POINTS

It follows that $P_{A B C D}=2 \cdot P_{\triangle A D C}$, hence $P_{\triangle A D C}=24 \mathrm{~cm}^{2}$. 3 POINTS

Therefore, $P_{\triangle A B C}=P_{A B C D}+P_{\triangle A D C}=48+24=72 \mathrm{~cm}^{2}$. 3 POINTS $\qquad$