## Task 2 - 320722

Let $A B C D$ be a square, with an area of $25 \mathrm{~cm}^{2}$. A point $E$ lies on the extension of the diagonal $A C$ beyond $C$, such that the segment $A E$ is twice as long as the segment $A C$.

Determine the area of the quadrilateral $A B E D$ under these conditions!
The triangles $A E D$ and $A C D$ share vertex $D$, which is opposite to sides $A E$ and $A C$ (which contains $A E$); thus, they have the same perpendicular height to these sides.

Therefore, and because $A E = 2 \cdot A C$, the area of $A E D$ is twice the area of $A C D$. Similarly (with $B$ instead of $D$): $A E B$ has twice the area of $A C B$.

Thus, we obtain: The area of $A B E D$ is twice the area of $A B C D$; hence, it is $50 \mathrm{~cm}^{2}$.