In a circle with a diameter of $10 \, \text{cm}$, there are two perpendicular chords $AB$ and $CD$, whose lengths are $9 \, \text{cm}$ and $8 \, \text{cm}$.

Calculate the distance from the intersection of lines $AB$ and $CD$ to the center of the circle.

(L. Hozová)
The center of the circle is denoted by $S$, the intersection of the chords is denoted by $P$, and the feet of the perpendiculars from $S$ to $A B$ and $C D$ are denoted by $Q$ and $R$ respectively:

![](https://cdn.mathpix.com/cropped/2024_04_17_9bcdd3b94ad4d4270464g-5.jpg?height=634&width=665&top_left_y=628&top_left_x=701)

The sought distance is the hypotenuse of the right triangle $S Q P$ or $S R P$, where each leg is identical to one of the legs of the right triangles $A Q S$ and $D R S$. We will determine all necessary values using the Pythagorean theorem repeatedly:

- In triangle $A Q S$, side $A Q$ is half of the chord $A B$ and side $S A$ is the radius of the circle, thus

$$
|S Q|^{2}=5^{2}-\left(\frac{9}{2}\right)^{2}=\frac{19}{4}\left(\mathrm{~cm}^{2}\right)
$$

- In triangle $D R S$, side $D R$ is half of the chord $C D$ and side $S D$ is the radius of the circle, thus

$$
|S R|^{2}=5^{2}-4^{2}=9\left(\mathrm{~cm}^{2}\right)
$$

- In triangle $S Q P$, we now know both legs, so

$$
|S P|^{2}=\frac{19}{4}+9=\frac{55}{4}\left(\mathrm{~cm}^{2}\right) .
$$

The distance from $P$ to $S$ is $\frac{\sqrt{55}}{2} \doteq 3.71(\mathrm{~cm})$.

Evaluation. 3 points for the analysis and auxiliary calculations; 3 points for the result and quality of the commentary.