A quadrilateral is inscribed in a circle of radius 10, with perpendicular diagonals measuring 12 and $10 \sqrt{3}$.

Find the sides of the quadrilateral.
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 $BD$. Then $BP = 5\sqrt{3}$,

$CK = 6$. Suppose point $P$ is between $M$ and $B$, and $M$ is between $K$ and $A$. Then $KM^2 = OP^2 = OB^2 - BP^2 = 25$, $MP^2 = OK^2 = OC^2 - KC^2 = 64$,

$BM = MP + PB = 8 + 5\sqrt{3}$, $AM = AK - KM = 1$, $AB = \sqrt{MB^2 + MA^2} = \sqrt{140 + 80\sqrt{3}} = \sqrt{80} + \sqrt{60}$.

The other sides are found similarly.

![](https://cdn.mathpix.com/cropped/2024_05_06_9e7ee8b57660b9721937g-43.jpg?height=480&width=463&top_left_y=1735&top_left_x=800)

## Answer

$4\sqrt{15} \pm 2\sqrt{5}, 4\sqrt{5} \pm 2\sqrt{15}$