Answer:

$$
4(2 \sqrt{13} \pm 1), \quad 4(8 \pm 2 \sqrt{13})
$$

## Solution

![](https://cdn.mathpix.com/cropped/2024_05_06_3e796a5bc771beb910ddg-2.jpg?height=411&width=419&top_left_y=1436&top_left_x=196)

Let $O$ be the center of the circumscribed circle of quadrilateral $ABCD$, $M$ be the intersection point of its diagonals, $K$ and $P$ be the midpoints of $AC$ and $BD$, $OK=8$, $OP=9$. Suppose that $M$ is between $P$ and $B$, and $K$ is between $M$ and $C$. Then $CK^2 = OC^2 - OK^2 = 225$, $DP^2 = OD^2 - OP^2 = 208 = 16 \cdot 13$, $CM = CK + KM = CK + OP = 24$, $BM = BP - MP = 4 \sqrt{13} - 8$, $BC = \sqrt{CM^2 + BM^2} = \sqrt{848 - 64 \sqrt{13}} = 4(2 \sqrt{13} - 1)$

The other sides are found similarly.
# Task 9.4

A quadrilateral is inscribed in a circle of radius 17, with diagonals that are perpendicular to each other and are 8 and 9 units away from the center of the circle. Find the sides of the quadrilateral.

Number of points 7