Task B-2.5. The area of square $A B C D$ is $256 \mathrm{~cm}^{2}$. On side $\overline{A D}$, we select point $E$, and side $\overline{D C}$ is extended beyond vertex $C$ to point $F$ such that angle $\measuredangle F B E$ is a right angle. Calculate the length of the extension $\overline{C F}$, if the area of triangle $E B F$ is $200 \mathrm{~cm}^{2}$.
## Solution.

![](https://cdn.mathpix.com/cropped/2024_05_30_8c954f607b7f1570b01bg-08.jpg?height=382&width=504&top_left_y=2162&top_left_x=775)

Let the side of the square be denoted by $a$, $|\overline{B E}|=x$, and $|\overline{B F}|=y$.

Notice that the angle $\measuredangle E B A=90^{\circ}-\measuredangle C B E=\measuredangle F B C$.

Then, by the SAS (Side-Angle-Side) congruence theorem, triangles $A B E$ and $C B F$ are congruent. Therefore, $x=y$.

(2 points)

Since the area of the square is $a^{2}=256 \mathrm{~cm}^{2}$, the length of the side $a$ is $a=16 \mathrm{~cm}$.

Similarly, from the area of triangle $E B F$ we get

$$
\begin{aligned}
\frac{1}{2} x y & =200 \\
x \cdot x & =400, x=20 \mathrm{~cm}
\end{aligned}
$$

Now, by the Pythagorean theorem,

$$
\begin{aligned}
|\overline{C F}|^{2} & =x^{2}-a^{2}=144 \\
|\overline{C F}| & =12 \mathrm{~cm}
\end{aligned}
$$