How can I calculate $AB$ in a quadrilateral figure $(ABCD)$ $AD=3cm$, $DC=4cm$, $BC=5cm$. The diagonals of this quad figure must also be perpendicular. Find $AB$.

Let $O$ be the intersecting point between the two diagonals $AC$ and $BD$.
Then by Pythagoras' Theorem we have the following identities:
$$\begin{cases}
AO^2 + DO^2 = AD^2 = 9 & (1)\\
DO^2 + CO^2 = CD^2 = 16 & (2)\\
CO^2 + BO^2 = BC^2 = 25 & (3)\\
BO^2 + AO^2 = AB^2 & (4)
 \end{cases}$$
Note that $(1) + (3) - (2) = (4)$ thus $AB^2 = 9+25-16 = 18\Rightarrow AB = 3\sqrt{2}$
