[ Auxiliary area. The area helps to solve the problem_] Heron's formula

The sides of the triangle are 13, 14, and 15. Find the radius of the circle that has its center on the middle side and touches the other two sides.
Let the center $O$ of the given circle be located on the side $AB$ of triangle $ABC$. The area of triangle $ABC$ is equal to the sum of the areas of triangles $AOC$ and $BOC$.

## Solution

Let $AC=13$, $AB=14$, $BC=15$, $O$ be the center of the given circle (with $O$ on side $AB$), $R$ be its radius, and $P$ and $Q$ be the points of tangency of the circle with sides $AC$ and $BC$ respectively. Using Heron's formula,

$$
S_{\triangle \mathrm{ABC}}=\sqrt{21(21-13)(21-14)(21-15)}=\sqrt{21 \cdot 8 \cdot 7 \cdot 6}=84
$$

Since $OP$ and $OQ$ are the heights of triangles $AOC$ and $BOC$, respectively, we have

$$
S_{\triangle \mathrm{ABC}}=S_{\triangle \mathrm{AOC}}+S_{\triangle \mathrm{BOC}}=\frac{1}{2} AC \cdot OP + \frac{1}{2} BC \cdot OQ = \frac{1}{2}(AC + BC) R = 14 R
$$

Therefore,

$$
R=\frac{S_{\triangle ABC}}{14}=\frac{84}{14}=6
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_b9b3326e3456e69a696eg-28.jpg?height=386&width=626&top_left_y=390&top_left_x=722)

Answer

6.