The center of a circle touching the legs $A C$ and $B C$ of a right triangle $A B C$ lies on the hypotenuse $A B$. Find the radius of the circle, if it is six times smaller than the sum of the legs, and the area of triangle $A B C$ is 27.
Let $O$ be the center of the circle, $M$ and $N$ be the points of tangency with the legs $A C$ and $B C$ respectively. Then $O M$ and $O N$ are the altitudes of triangles $A O C$ and $B O C$. Let $B C=a, A C=b, O M=O N=r$. According to the problem, $a+b=6 r$. Then

$$
27=S_{\triangle A B C}=S_{\triangle A O C}+S_{\triangle B O C}=\frac{1}{2} a r+\frac{1}{2} b r=\frac{1}{2}(a+b) r=3 r 2,
$$

from which we find that

$$
r 2=\frac{27}{3}=9, r=3
$$

## Answer

3.00