A cyclic quadrilateral has consecutive sides of 36, 91, 315, and 260 units. Let's calculate the radius of the circumscribed circle and the radius of the inscribed circle!
In the convex cyclic quadrilateral $ABCD$, let $AB=a=36, BC=b=91, CD=c=315, DA=d=260$ units, and $\angle BAD = \alpha$.

![](https://cdn.mathpix.com/cropped/2024_05_02_4080b5279d05cc4b7975g-1.jpg?height=548&width=873&top_left_y=257&top_left_x=623)

Thus, $\angle DCB = 180^\circ - \alpha$ and

$$
\begin{gathered}
BD^2 = e^2 = a^2 + d^2 - 2ad \cos \alpha = b^2 + c^2 + 2bc \cos \alpha \\
\cos \alpha = \frac{(a^2 + d^2) - (b^2 + c^2)}{2(ad + bc)} = -\frac{33}{65}
\end{gathered}
$$

Furthermore,

$$
e = 280, \quad \sin \alpha = \frac{56}{65}
$$

and from these, the radius of the circumscribed circle is

$$
r = \frac{e}{2 \sin \alpha} = 162.5 \text{ units. }
$$

The sum of the opposite sides of our quadrilateral is equal: $36 + 315 = 91 + 260 = s$, which is half the perimeter, so there exists an inscribed circle. To find its radius, the area of the quadrilateral is

$$
t = \frac{1}{2} ad \sin \alpha + \frac{1}{2} bc \sin \alpha = 16380 \text{ units, }
$$

thus the radius of the inscribed circle is

$$
\rho = \frac{t}{s} = \frac{140}{3} \text{ units. }
$$