Example 6. The center of square $ABCD$ 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 $P B=$ $\qquad$ . (1989, National Junior High School League)
Analysis: The answer is $P B=42 \mathrm{~cm}$. How do we get it?
Connect $O A, O B$. It is easy to know that $O, P, A, B$ are concyclic, so $\angle A P B=\angle A O B=90^{\circ}$.

Therefore, $P A^{2}+P B^{2}=A B^{2}=1989$. Since $P A: P B=5: 14$, we can find $P B$.