8,9

A circle inscribed in a triangle touches one of the sides, dividing it into segments of 3 and 4, and the angle opposite this side is $120^{\circ}$. Find the area of the triangle.
Let $x$ be the distance from the vertex of the angle of $120^{\circ}$ to the nearest point of tangency, and apply the Law of Cosines. We obtain the equation $x^2 + 7x - 4 = 0$, from which we find that $x = \frac{\sqrt{65} - 7}{2}$.

Let $p$ be the semiperimeter of the triangle, $r$ be the radius of the inscribed circle, and $S$ be the area. Then

$$
p = 3 + 4 + \frac{\sqrt{65} - 7}{2} = \frac{\sqrt{65} + 7}{2}, \quad r = x \tan 60^{\circ} = x \sqrt{3} = \frac{(\sqrt{65} - 7) \sqrt{3}}{2}
$$

Therefore,

$$
S = p r = \frac{\sqrt{65} + 7}{2} \cdot \frac{(\sqrt{65} - 7) \sqrt{3}}{2} = 4 \sqrt{3}
$$

## Answer

Problem