The triangle $\triangle A B C$ is equilateral and the point $P$ is such that $P A=3 \text{~cm}$, $P B=4 \text{~cm}$ and $P C=5 \text{~cm}$. Calculate the length of the sides of the triangle $\triangle A B C$.

![](https://cdn.mathpix.com/cropped/2024_05_01_30c9a294a58a6e4b8190g-026.jpg?height=448&width=528&top_left_y=593&top_left_x=1301)
Since the triangle $\triangle A B C$ is equilateral, its angles are all equal to $60^{\circ}$. On the side $C B$ of this triangle, we construct a new triangle $\triangle C B P^{\prime}$, congruent to the triangle $\triangle A B P$, such that $P \widehat{A} B=P^{\prime} \widehat{C} B$ and $A \widehat{B} P=C \widehat{B} P^{\prime}$ (by rotating the triangle $\triangle A B P$ clockwise by $60^{\circ}$ around the point $B$).

![](https://cdn.mathpix.com/cropped/2024_05_01_30c9a294a58a6e4b8190g-091.jpg?height=631&width=762&top_left_y=1809&top_left_x=727)

Note that the angle $P \widehat{B} P^{\prime}$ is congruent to the angle $A \widehat{B} C$, that is, it measures $60^{\circ}$. Thus, if we draw the segment $P P^{\prime}$, we have that the triangle $\triangle P B P^{\prime}$, which is already isosceles, since $P B=B P^{\prime}=4 \mathrm{~cm}$, is in fact equilateral, and consequently, we have that $P P^{\prime}=4 \mathrm{~cm}$.

![](https://cdn.mathpix.com/cropped/2024_05_01_30c9a294a58a6e4b8190g-092.jpg?height=600&width=720&top_left_y=251&top_left_x=702)

Denoting the angle $P \widehat{P^{\prime}} C$ by $a$, we apply the Law of Cosines to the triangle $\triangle C P P^{\prime}$ and obtain

$$
5^{2}=3^{2}+4^{2}-2 \cdot 3 \cdot 4 \cdot \cos a
$$

that is, $25=25-24 \cos a$. It follows that $\cos a=0$ and, therefore, $a=90^{\circ}$. Thus, we establish $C \widehat{P^{\prime} B} B=a+60^{\circ}=90^{\circ}+60^{\circ}=150^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_05_01_30c9a294a58a6e4b8190g-092.jpg?height=614&width=714&top_left_y=1206&top_left_x=705)

Now, denoting the side of the triangle $\triangle A B C$ by $l$, we apply the Law of Cosines to the triangle $\triangle C B P^{\prime}$ and obtain

$$
l^{2}=3^{2}+4^{2}-2 \cdot 3 \cdot 4 \cdot \cos 150^{\circ}=25+12 \sqrt{3}
$$

It follows that $\sqrt{25+12 \sqrt{3}} \mathrm{~cm}$ is the length of the sides of the equilateral triangle $\triangle A B C$.