Isosceles trapezoid $A B C D$ with bases $A B$ and $C D$ has a point $P$ on $A B$ with $A P=11, B P=27$, $C D=34$, and $\angle C P D=90^{\circ}$. Compute the height of isosceles trapezoid $A B C D$.
Solution: Drop projections of $A, P, B$ onto $C D$ to get $A^{\prime}, P^{\prime}, B^{\prime}$. Since $A^{\prime} B^{\prime}=38$ and $C D=34$, we get that $D A^{\prime}=C B^{\prime}=2$. Thus, $P^{\prime} D=9$ and $P^{\prime} C=25$. Hence, the answer is $P P^{\prime}=\sqrt{P^{\prime} D \cdot P^{\prime} C}=$ 15. $\fbox{15}$.