### 10.417

In a right triangle $ABC\left(\angle C=90^{\circ}\right)$, a circle is inscribed, touching its sides at points $A_{1}, B_{1}, C_{1}$. Find the ratio of the area of triangle $ABC$ to the area of triangle $A_{1} B_{1} C_{1}$, if $AC=4 \text{ cm}, BC=3 \text{ cm}$.
## Solution.

![](https://cdn.mathpix.com/cropped/2024_05_21_f024bf2ff7725246f3bfg-252.jpg?height=256&width=373&top_left_y=499&top_left_x=87)
Let $O$ be the center of the circle with radius $r$ inscribed in the given $\triangle ABC$.
Then $AB=5, r=\frac{3+4-5}{2}=1$ (Theorem 1).
$S_{A_{1} B_{1} C_{1}}=S_{O A_{1} B_{1}}+S_{O A_{1} C_{1}}+S_{O B_{1} C_{1}}=$
$=\frac{1}{2} r^{2} \sin \angle A_{1} O B_{1}+\frac{1}{2} r^{2} \sin \angle A_{1} O C_{1}+$

$+\frac{1}{2} r^{2} \sin \angle C_{1} O B_{1}=\frac{r^{2}}{2}(\sin \angle C+\sin \angle A+\sin \angle B)=\frac{1}{2}\left(1+\frac{3}{5}+\frac{4}{5}\right)=\frac{6}{5} \Rightarrow \frac{S_{A B C}}{S_{A_{1} B_{1} C_{1}}}=5$.

## Answer: 5.