## Task 3 - 170933

In a triangle $ABC$, let $AC=b=13 \text{~cm}$ and $BC=a=15 \text{~cm}$. The perpendicular from $C$ to the line through $A$ and $B$ is $CD$, and it is given that $CD=h_{c}=12 \text{~cm}$.

Determine the area $I$ for all triangles $ABC$ that meet these conditions!
The triangle $\triangle A C D$ is right-angled with the right angle at $D$. According to the Pythagorean theorem, we have $|A D|^{2}+|C D|^{2}=|A C|^{2}$, so because $\sqrt{13^{2}-12^{2}}=\sqrt{169-144}=\sqrt{25}=5$, it follows that $|A D|=5 \mathrm{~cm}$.

In the same way, in the right-angled triangle $\triangle B C D$, we get $\sqrt{15^{2}-12^{2}}=\sqrt{225-144}=\sqrt{81}=9$, so $|B D|=9 \mathrm{~cm}$.

If $A$ and $B$ lie on the same side of $D$ on the line through $A$ and $B$ (meaning that $\triangle A B C$ is an obtuse triangle, with the foot of the altitude $D$ lying outside), then $c=|A B|=|B D|-|A D|=4 \mathrm{~cm}$, and thus $I=\frac{1}{2} c \cdot h_{c}=24 \mathrm{~cm}^{2}$.

If, on the other hand, $A$ and $B$ lie on different sides of $D$ on the line through $A$ and $B$ (i.e., $D$ is between them), then $c=|A B|=|A D|+|B D|=14 \mathrm{~cm}$, and thus $I=84 \mathrm{~cm}^{2}$.