5. As shown in the figure, the center of square $A B C D$ is $O$, and its area is $1989 \mathrm{~cm}^{2}, P$ is a point inside the square, and $\angle O P B=45^{\circ}, P A: P B=5: 14$, then $P B=$ $\qquad$ $\mathrm{cm}$.
5. 42

Solution: Connect $O A, O B$. Since $\angle B A O=\angle B P O=45^{\circ}$, we know that points $A, B, O, P$ are concyclic, thus $\angle A P B=\angle A O B=90^{\circ}$. Let $P A=5 x, P B=14 x$, then $(5 x)^{2}+(14 x)^{2}=1989$.
Solving for $x$ gives $x=3$, so $P B=42$.