10.419 A circle with its center on side $AB$ of triangle $ABC$ touches the other two sides of the triangle. Find the area of the circle if $a=13$ cm, $b=14$ cm, $c=15 \mathbf{cm}$, where $a, b$ and $c$ are the lengths of the sides of the triangle.
## Solution.

Let $O$ be the center of the circle with radius $r$, touching

![](https://cdn.mathpix.com/cropped/2024_05_21_f024bf2ff7725246f3bfg-253.jpg?height=273&width=314&top_left_y=555&top_left_x=196)
sides $A C$ and $B C$ of the given $\triangle A B C$ at points $K$ and $M$. By the cosine rule, $\cos \angle C=\frac{a^{2}+b^{2}-c^{2}}{2 a b}=\frac{5}{13}$, $\sin \angle C=\frac{12}{13} \Rightarrow S_{A B C}=\frac{a b \sin \angle C}{2}=84$.
On the other hand, $S_{A B C}=S_{A O C}+S_{B O C}=$ $=\frac{r \cdot b}{2}+\frac{r \cdot a}{2}=r \cdot \frac{27}{2} \Rightarrow r=\frac{56}{9}, \pi r^{2}=\frac{3136}{81} \pi$.

Answer: $\frac{3136}{81} \pi$.