# Problem 5

The bisector of a right triangle divides a leg into segments of lengths 25 cm and 24 cm. Find the length of the circumscribed circle around this triangle.
Answer: The length of the circumscribed circle is $c=175 \pi$.

## Solution

![](https://cdn.mathpix.com/cropped/2024_05_06_bea8bd31fc71aedf7f6dg-4.jpg?height=1245&width=876&top_left_y=622&top_left_x=293)

Drop a perpendicular from point M to the hypotenuse. Triangles $A C M$ and $A M N$ are equal by hypotenuse and acute angle. Then $M N=24$ cm, and by the Pythagorean theorem, $N B=7$ cm. From the similarity of right triangles $A B C$ and $M N B$, we get $\frac{A B}{B M}=\frac{B C}{B N} ; \quad \frac{A B}{25}=\frac{49}{7}$. Therefore, the hypotenuse is 175 cm. Since the hypotenuse is the diameter of the circumscribed circle, $c=175 \pi$.