Right-angled triangle $ABC$ has legs $|AB|=5 \text{ cm}$ and $|BC|=12 \text{ cm}$. The vertices of triangle $ABC$ are mapped such that

- point $A$ is mapped to point $A^{\prime}$ symmetrically with respect to the angle bisector of $\angle ABC$,
- point $B$ is mapped to point $B^{\prime}$ symmetrically with respect to the midpoint of segment $BC$,
- point $C$ is mapped to point $C^{\prime}$ symmetrically with respect to the perpendicular bisector of segment $AB$.

Determine the area of triangle $A^{\prime} B^{\prime} C^{\prime}$.

(M. Dillingerová)
The rays of angle $ABC$ are symmetric with respect to its axis. Therefore, point $A^{\prime}$ lies on the ray $BC$, and $\left|B A^{\prime}\right|=|B A|=5 \mathrm{~cm}$.

The endpoints of segment $BC$ are symmetric with respect to its midpoint, so $B^{\prime}=C$.

Points $C$ and $C^{\prime}$ are symmetric with respect to the axis of segment $AB$, so segment $CC^{\prime}$ is perpendicular to this line. Additionally, triangle $ABC$ is a right triangle with a right angle at vertex $B$, so points $A$, $B$, $C$, and $C^{\prime}$ form the vertices of a rectangle.

![](https://cdn.mathpix.com/cropped/2024_04_17_e1de9eaadf2ab4e99947g-2.jpg?height=822&width=514&top_left_y=1685&top_left_x=771)

Triangle $A^{\prime} B^{\prime} C^{\prime}$ is thus a right triangle with a right angle at vertex $B^{\prime}$. Its legs have lengths

$$
\begin{aligned}
& \left|A^{\prime} B^{\prime}\right|=\left|B B^{\prime}\right|-\left|B A^{\prime}\right|=12-5=7(\mathrm{~cm}), \\
& \left|B^{\prime} C^{\prime}\right|=|A B|=5(\mathrm{~cm}) .
\end{aligned}
$$

The area of triangle $A^{\prime} B^{\prime} C^{\prime}$ is $\frac{1}{2} \cdot 7 \cdot 5=17.5\left(\mathrm{~cm}^{2}\right)$.

Scoring suggestion. 1 point for accurately determining the positions of points $A^{\prime}, B^{\prime}, C^{\prime} ; 2$ points for recognizing that triangle $A^{\prime} B^{\prime} C^{\prime}$ is a right triangle; 1 point for the lengths of the legs and the area.