## Task B-2.2.

Starting from the shortest side of the triangle, each subsequent side is $4 \mathrm{~cm}$ longer than the previous one. If the measure of one of the angles of this triangle is $120^{\circ}$, calculate the radius of the circle inscribed in the triangle.
## Solution.

Let the sides of the triangle be $a, b$, and $c$. From the conditions of the problem, we have $b=a+4, c=a+8$, and the angle $\gamma=120^{\circ}$ is opposite the largest side.

Applying the cosine rule, we get

$$
\begin{gathered}
c^{2}=a^{2}+b^{2}-2 a b \cos \gamma \\
(a+8)^{2}=a^{2}+(a+4)^{2}-2 a(a+4) \cos 120^{\circ}
\end{gathered}
$$

After simplification, we obtain the quadratic equation $a^{2}-2 a-24=0$. 1 point

The positive solution to this equation is $a=6$.

Finally, the sides of the triangle are $6 \mathrm{~cm}, 10 \mathrm{~cm}$, and $14 \mathrm{~cm}$.

The area of the triangle is calculated using the formula $P=\frac{a b}{2} \sin \measuredangle(a, b)$:

$$
P=\frac{6 \cdot 10}{2} \sin 120^{\circ}=30 \cdot \frac{\sqrt{3}}{2}=15 \sqrt{3} \mathrm{~cm}^{2}
$$

Then, $r=\frac{P}{s}=\frac{15 \sqrt{3}}{15}=\sqrt{3} \mathrm{~cm}$.