In the circle $k$ with center $S$ and radius $52 \, \text{mm}$, two perpendicular chords $A B$ and $C D$ are given. Their intersection $X$ is $25 \, \text{mm}$ away from the center $S$. How long is the chord $C D$, if the length of the chord $A B$ is $96 \, \text{mm}$?

(L. Hozová)
The midpoints of chords $AB$ and $CD$ are denoted as $E$ and $F$, see the figure. Triangles $AES$, $EXS$, and $SFD$ are right-angled, $|AE|=\frac{1}{2}|AB|=48 \text{ mm}$, $|SX|=25 \text{ mm}$, and $|SA|=|SD|=$ $=52 \text{ mm}$.

![](https://cdn.mathpix.com/cropped/2024_04_17_8faaf4d15d1fed209efag-3.jpg?height=859&width=857&top_left_y=176&top_left_x=605)

According to the Pythagorean theorem in triangle $AES$, we get

$$
|SE|^2 = |SA|^2 - |AE|^2 = 52^2 - 48^2 = 2704 - 2304 = 400 \text{ mm}^2.
$$

According to the Pythagorean theorem in triangle $EXS$, we get

$$
|EX|^2 = |SX|^2 - |SE|^2 = 25^2 - 400 = 625 - 400 = 225 \text{ mm}^2.
$$

Since $EXFS$ is a rectangle, $|SF| = |EX|$; according to the Pythagorean theorem in triangle $SFD$, we get

$$
|FD|^2 = |SD|^2 - |SF|^2 = 52^2 - 225 = 2704 - 225 = 2479 \text{ mm}^2.
$$

From this, we express $|FD| \approx 49.79$, or $|FD| \approx 50 \text{ mm}$ if working with whole mm precision. The chord $CD$ is thus approximately $100 \text{ mm}$ long.

Evaluation. 1 point for determining $|AE|$; 1 point for determining $|SE|$; 1 point for determining $|EX|$; 1 point for determining $|SF|$; 1 point for determining $|FD|$; 1 point for determining $|CD|$.