7. On the side $B C$ of triangle $A B C$, a point $K$ is marked.

It is known that $\angle B+\angle C=\angle A K B, \quad A K=4, \quad B K=9, \quad K C=3$.

Find the area of the circle inscribed in triangle $A B C$.
# Solution:

$\angle B+\angle C=\angle A K B, \quad \angle A K C=\angle A, \quad \angle K A C=\angle B$,

$\triangle A B C \approx \triangle A K C \Rightarrow$

$\frac{A B}{A K}=\frac{A C}{K C}=\frac{B C}{A C} \Rightarrow \frac{A B}{4}=\frac{A C}{3}=\frac{12}{A C} \Rightarrow$

$A C=6, \quad A B=8$.

![](https://cdn.mathpix.com/cropped/2024_05_06_797a639488ab43e0ad60g-26.jpg?height=417&width=591&top_left_y=255&top_left_x=1301)

$S_{A B C}=\sqrt{p(p-A B)(p-B C)(p-A C)}, \quad p=\frac{1}{2}(A B+B C+A C)=\frac{1}{2}(8+12+6)=13$,

$S_{A B C}=\sqrt{13(13-8)(13-12)(13-6)}=\sqrt{13 \cdot 5 \cdot 7}=\sqrt{35 \cdot 13} ; \quad r_{\varepsilon n}=\frac{S_{A B C}}{p}=\frac{\sqrt{35}}{\sqrt{13}} ; \quad S_{\varepsilon n k p}=\frac{35}{13} \pi$.

Answer: $\frac{35}{13} \pi$.