5. In the figure, rectangle $A B C D$ and triangle $A B F$ are shown. What is the area of triangle $A B F$ if the following conditions are met: $|\angle A F D|=|\angle F B C|,|F D|=6 \text{ cm}$ and $|F C|=2 \text{ cm}$?

![](https://cdn.mathpix.com/cropped/2024_06_03_1be9b24bbc58be2cc921g-1.jpg?height=426&width=805&top_left_y=1980&top_left_x=637)

## Problems worth 10 points:
First method:

![](https://cdn.mathpix.com/cropped/2024_05_30_64eca78245e6b7f11e3dg-3.jpg?height=351&width=663&top_left_y=1015&top_left_x=294)

Let $|A D|=x$.

Triangles $\triangle A F D$ and $\triangle F B C$ are similar right triangles by the $\mathrm{KK}$ theorem \quad 1 POINT

$\left(|\angle A F D|=|\angle F B C|\right.$ and both have angles of $90^{\circ}$ ).

1 POINT

From the similarity of the triangles, the ratio equality holds: $\frac{6}{x}=\frac{x}{2}$.

1 POINT

Furthermore, $x^{2}=12$, so $x=2 \sqrt{3} \mathrm{~cm}$.

1 POINT

Let $E$ be the foot of the perpendicular from vertex $F$ to side $\overline{A B}$.

Then $|E F|=2 \sqrt{3} \mathrm{~cm}$,

1 POINT

so the area of triangle $\triangle A B F$ is $P=\frac{8 \cdot 2 \sqrt{3}}{2} \mathrm{~cm}^{2}=8 \sqrt{3} \mathrm{~cm}^{2}$.

1 POINT

.TOTAL 6 POINTS

Note: If the process of partial root extraction is not derived, 1 POINT should be deducted.