, 2025

Let $ABCD$ be the trapezoid with $\overline{AB} \parallel \overline{CD}$ and $AB = s < r = CD$. Let $K, L, M$, and $N$ be the points at which the circle is tangent to $\overline{AB}$, $\overline{BC}$, $\overline{CD}$, and $\overline{DA}$, respectively. Because $KM$ is a diameter of the circle perpendicular to $\overline{AB}$ and $\overline{CD}$, the trapezoid has height $6$. Therefore
$$72 = \text{Area}(ABCD) = 6 \cdot \frac{r+s}{2},$$
so $r + s = 24$.

By the Equal Tangents Theorem $AK = AN$, $BK = BL$, $CL = CM$, and $DM = DN$, so $AB + CD = AD + BC$. Hence $2AD = AD + BC = AB + CD = r + s$, implying $AD = 12$. Let $P$ be the foot of the perpendicular from $A$ to $\overline{CD}$. Then $DP = \frac{r-s}{2}$, and by the Pythagorean Theorem $DP^2 = AD^2 - AP^2$, so
$$\left( \frac{r-s}{2} \right)^2 = 12^2 - 6^2 = 108,$$
from which $(r-s)^2 = 432$. Thus
$$r^2 + s^2 = \frac{(r+s)^2 + (r-s)^2}{2} = \frac{24^2 + 432}{2} = 504.$$