3. In triangle $ABC$, the lengths of the sides $|AB|=48 \, \text{cm}$ and $|AC|=30 \, \text{cm}$ are known, as well as the median $|CP|=18 \, \text{cm}$. Calculate the perimeter and area of triangle $ABC$.
3. 

![](https://cdn.mathpix.com/cropped/2024_05_30_e13d523bf436e84e6a35g-19.jpg?height=323&width=391&top_left_y=1118&top_left_x=273)

We are given triangle $A P C$. Its sides have lengths $|A P|=24 \mathrm{~cm}, |P C|=18 \mathrm{~cm}$, and $|C A|=30 \mathrm{~cm}$.

$1 \mathrm{POINT}$

The sum of the areas of the squares constructed on the two shorter sides is

$18^{2}+24^{2}=324+576=900$

$1 \mathrm{POINT}$

The area of the square on its longest side is $30^{2}=900$.

$1 \mathrm{POINT}$

Since $18^{2}+24^{2}=30^{2}$, by the converse of the Pythagorean theorem, we conclude that triangle $A P C$ is a right triangle with a right angle at vertex $P$.

2 POINTS

![](https://cdn.mathpix.com/cropped/2024_05_30_e13d523bf436e84e6a35g-19.jpg?height=319&width=376&top_left_y=2065&top_left_x=286)

Triangles $\triangle A P C$ and $\triangle B P C$ are right triangles and share the leg $\overline{C P}$. Given that $|A P|=|B P|$, it follows that $|A C|=|B C|=30 \mathrm{~cm}$, i.e., triangle $A B C$ is isosceles with base $\overline{A B}$.

The perimeter of triangle $A B C$ is $O=48+2 \cdot 30=108 \mathrm{~cm}$.

$1 \mathrm{POINT}$

Since $\overline{C P} \perp \overline{A B}$, the median $\overline{C P}$ is also the altitude from vertex $C$.

The area of triangle $A B C$ is $P_{A B C}=\frac{|A B| \cdot|C P|}{2}=\frac{48 \cdot 18}{2}=432 \mathrm{~cm}^{2}$. $\qquad$
TOTAL 10 POINTS