1.2. Inside the rectangle $A B C D$, whose sides are $A B=C D=15$ and $B C=A D=10$, there is a point $P$ such that $A P=9, B P=12$. Find $C P$.
1.2. $A B=15, B P=12, A P=9$. Therefore, $A B P$ is a right triangle. Drop the heights $P S$ and $P T$ onto $A B$ and $B C$ respectively. From similarity, we get $P S=A P \cdot B P / A B=36 / 5, P T=B P^{2} / A B=48 / 5$. Then $C P^{2}=$ $(B C-P S)^{2}+P T^{2}=2500 / 25=100, C P=10$.