Eight circles of radius $r$ located in a right triangle $ABC$ (angle $C$ is right) as shown in figure (each of the circles touches the respactive sides of the triangle and the other circles). Find the radius of the inscribed circle of triangle $ABC$.
[img]https://cdn.artofproblemsolving.com/attachments/4/7/1b1cd7d6bc7f5004b8e94468d723ed16e9a039.png[/img]
1. **Identify the relationship between the triangles:**
   - Given that there are 8 circles of radius \( r \) located in a right triangle \( \triangle ABC \) with a right angle at \( C \).
   - The centers of these circles form a smaller right triangle \( \triangle DEF \) similar to \( \triangle ABC \).

2. **Determine the sides of \( \triangle DEF \):**
   - Since each circle touches the respective sides of the triangle and the other circles, the sides of \( \triangle DEF \) are \( 6r \), \( 8r \), and \( 10r \) (using the Pythagorean theorem for the right triangle).

3. **Calculate the radius of the incircle of \( \triangle DEF \):**
   - The formula for the radius \( r_{\text{in}} \) of the incircle of a right triangle with legs \( a \) and \( b \) and hypotenuse \( c \) is:
     \[
     r_{\text{in}} = \frac{a + b - c}{2}
     \]
   - For \( \triangle DEF \) with sides \( 6r \), \( 8r \), and \( 10r \):
     \[
     r_{\text{in}} = \frac{6r + 8r - 10r}{2} = \frac{4r}{2} = 2r
     \]

4. **Relate the incircle radius of \( \triangle DEF \) to \( \triangle ABC \):**
   - Since \( \triangle DEF \sim \triangle ABC \), the ratio of their corresponding sides is the same.
   - The sides of \( \triangle ABC \) are \( 9r \), \( 12r \), and \( 15r \) (since \( \triangle ABC \) is scaled up by a factor of \( \frac{3}{2} \) from \( \triangle DEF \)).

5. **Calculate the radius of the incircle of \( \triangle ABC \):**
   - Using the same formula for the radius of the incircle:
     \[
     r_{\text{in}} = \frac{a + b - c}{2}
     \]
   - For \( \triangle ABC \) with sides \( 9r \), \( 12r \), and \( 15r \):
     \[
     r_{\text{in}} = \frac{9r + 12r - 15r}{2} = \frac{6r}{2} = 3r
     \]

Conclusion:
\[
\boxed{3r}
\]