Gara di Febbraio

The answer is $(\mathbf{D})$. Let, as in the figure, $K$ be the meeting point of the diagonals, $x_{1}, x_{2}$ the lengths of the segments $AK, KC$ and $y_{1}, y_{2}$ those of the segments $BK, KD$ respectively. Since all the angles at $K$ are right angles, by the Pythagorean theorem we have
$$\left\{ \begin{array}{l} x_{1}^{2}+y_{1}^{2}=AB^{2}=100^{2} \\ y_{1}^{2}+x_{2}^{2}=BC^{2}=120^{2} \\ x_{2}^{2}+y_{2}^{2}=CD^{2}=75^{2} \\ y_{2}^{2}+x_{1}^{2}=DA^{2} \end{array} \right.$$

By adding the first and third equations and subtracting the second, we then obtain
$$DA^{2}=y_{2}^{2}+x_{1}^{2}=(x_{1}^{2}+y_{1}^{2})+(x_{2}^{2}+y_{2}^{2})-(y_{1}^{2}+x_{2}^{2})=100^{2}+75^{2}-120^{2}.$$

A simple calculation now gives the answer: $DA^{2}=100^{2}+75^{2}-120^{2}=5^{2} \cdot (20^{2}+15^{2}-24^{2})=5^{2} \cdot (400+225-576)=5^{2} \cdot 49=5^{2} \cdot 7^{2}=35^{2}$. Finally, observe that the same answer is obtained even if the quadrilateral $ABCD$ is not convex: in that case, the point $D$ in the figure is replaced by the point $D'$, its reflection with respect to $K$, and the same reasoning can be applied.