The vertices of the triangle are connected to the center of the inscribed circle. The segments divide the area of the triangle into three parts, equal to 28, 60, and 80. Find the sides of the triangle.

#
Apply Heron's formula.

## Solution

Let $a, b$ and $с$ be the sides of the triangle, which are the bases of triangles with areas 28, 60, and 80, respectively, and $r$ be the radius of the inscribed circle of the triangle. Then

$$
\frac{1}{2} a r=28, \frac{1}{2} b r=60, \frac{1}{2} c r=80
$$

From these equations, express $a, b$ and $с$ in terms of $r$:

$$
a=\frac{56}{r}, b=\frac{120}{r}, c=\frac{160}{r}
$$

Using Heron's formula, express the area of the given triangle in terms of $r$:

$$
28+60+80=\sqrt{\frac{168}{r} \cdot \frac{112}{r} \cdot \frac{48}{r} \cdot \frac{8}{r}}
$$

From the obtained equation, we find that $r=4$. Therefore, $a=14, b=30, c=40$.

![](https://cdn.mathpix.com/cropped/2024_05_06_16ab69fab1743849dba9g-32.jpg?height=289&width=1033&top_left_y=819&top_left_x=518)

Answer

14, 30, and 40.