[Inscribed, circumscribed, and exscribed circles; their radii $]$
[ Auxiliary area. The area helps to solve the task_ ]

In an isosceles triangle, the base is 30, and the lateral side is 39. Find the radius of the inscribed circle.
The radius of the circle inscribed in a triangle is equal to the area of the triangle divided by its semi-perimeter.

## Solution

First method.

Let's find the height of the triangle dropped to the base:

$$
h^{2}=39^{2}-15^{2}=24 \cdot 54=36^{2}, h=36
$$

The desired radius is equal to the area of the triangle divided by its semi-perimeter:

$$
r=\frac{S}{p}=\frac{15 \cdot 36}{18}=10
$$

Second method.

Since the triangle is isosceles, the center $O$ of its inscribed circle lies on the height $A M$ dropped to the base $B C$. Let $K$ be the point of tangency of the inscribed circle with the lateral side $A C$. Then

$$
K C=C M=15, A K=39-15=24, A M=36 .
$$

The radius $O K$ of the inscribed circle will be found from the similarity of triangles $A O K$ and $A C M$.

$$
\text { Third method. }
$$

By the property of the angle bisector of a triangle $A O: O M=A B: B M$. Therefore,

$$
r=O M=\frac{15 \cdot 36}{15+39}=10
$$

## Answer

10.