4. The quadrilateral $A B C D$ has perpendicular diagonals. It is also known that $A B=100, B C=120$, $C D=75$. Determine the length of $A D$.
(A) 30
(B) $24 \sqrt{2}$
(C) $20 \sqrt{3}$
(D) 35
(E) $\frac{125}{2}$
4. The answer is $(\mathbf{D})$. Let, as in the figure, $K$ be the intersection point of the diagonals, $x_{1}, x_{2}$ the lengths of the segments $A K, K C$ and $y_{1}, y_{2}$ those of the segments $B K, K D$ respectively. Since all angles at $K$ are right angles, by the Pythagorean theorem we have

$$
\left\{\begin{array}{l}
x_{1}^{2}+y_{1}^{2}=A B^{2}=100^{2} \\
y_{1}^{2}+x_{2}^{2}=B C^{2}=120^{2} \\
x_{2}^{2}+y_{2}^{2}=C D^{2}=75^{2} \\
y_{2}^{2}+x_{1}^{2}=D A^{2}
\end{array}\right.
$$

By adding the first and third equations and subtracting the second, we obtain

$D A^{2}=y_{2}^{2}+x_{1}^{2}=\left(x_{1}^{2}+y_{1}^{2}\right)+\left(x_{2}^{2}+y_{2}^{2}\right)-\left(y_{1}^{2}+x_{2}^{2}\right)=100^{2}+75^{2}-120^{2}$.

![](https://cdn.mathpix.com/cropped/2024_04_17_31098356af162238e164g-11.jpg?height=614&width=511&top_left_y=1755&top_left_x=1361)

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