33. In the following diagram, $A B C D$ is a square, and $E$ is the center of the square $A B C D . P$ is a point on a semi-circle with diameter $A B$. $Q$ is a point on a semi-circle with diameter $A D$. Moreover, $Q, A$ and $P$ are collinear (that is, they are on the same line). Suppose $Q A=14$ $\mathrm{cm}, A P=46 \mathrm{~cm}$, and $A E=x \mathrm{~cm}$. Find the value of $x$.
33. Ans: 34 .
Join $Q D$ and $P B$, it is easy to show that right $\triangle D Q A$ is congruent to right $\triangle A P B$. Thus $P B=Q A=14$.
Apply Pythagoras' Theorem to $\triangle A P B$,
$$
(\sqrt{2} x)^{2}=46^{2}+14^{2} .
$$

Solve, $x=34$.