# Solution:

According to the problem: $K N=a=2$ - the distance from point $K$ to line $B C$, $K M=b=12$ - the distance from point $K$ to line $A B$, $r=5$ - the radius of the circle inscribed in triangle $A B C$. Let $O$ be the intersection point of the diagonals of the rhombus, and $E$ be the center of the circle inscribed in triangle $A B C$.

Let $B O=h, O Q=r_{1}$ - the radius of the circle inscribed in rhombus $A B C D$, and $A P=h_{1}$ - the height of triangle $A B C$ drawn from vertex $A$ to side $B C$.

1) $h_{1}=a+b=14$
2) $S_{A B C}=\frac{1}{2} A C \cdot h, S_{A B C}=\frac{1}{2} B C \cdot h_{1}, S_{A B C}=p r$
3) $B C \cdot h_{1}=(A C+2 B C) \cdot r \Rightarrow$

$$
14 B C=5(A C+2 B C) \Rightarrow 4 B C=5 A C
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_18575ac96ab43cb21cd9g-05.jpg?height=1302&width=585&top_left_y=223&top_left_x=1318)
4) $A C \cdot h=B C \cdot h_{1} \Rightarrow A C \cdot h=1.25 A C \cdot h_{1} \Rightarrow$

$$
h=17.5
$$

5) $B E=h-r=12.5, \sin (\angle O B C)=\frac{E S}{B E}=\frac{r}{B E}=\frac{5}{12.5}=\frac{2}{5}, \cos (\angle O B C)=\frac{\sqrt{21}}{5}$,

$$
B C=\frac{h}{\cos (\angle O B C)}=\frac{17.5 \cdot 5}{\sqrt{21}}=\frac{25 \sqrt{21}}{6} \text {. }
$$

6) $r_{1}=0.5 h_{1}=7$.

Answer: $\frac{25 \sqrt{21}}{6}, 7$.
7. On the diagonal $AC$ of the rhombus $ABCD$, a point $K$ is chosen, which is at a distance of 12 and 2 from the lines $AB$ and $BC$ respectively. The radius of the circle inscribed in triangle $ABC$ is 5. Find the side of the rhombus $ABCD$ and the radius of the circle inscribed in this rhombus.

(12 points)

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