9. In square $A B C D, P$ and $Q$ are points on sides $C D$ and $B C$, respectively, such that $\angle A P Q=90^{\circ}$. If $A P=4$ and $P Q=3$, find the area of $A B C D$.
Answer: $\frac{256}{17}$
Solution: Note that triangles $A D P$ and $P C Q$ are similar, so $A D / P C=A P / P Q=4 / 3$. Let $A D=4 x$ and $P C=3 x$. Since $A B C D$ is a square, $P D=x$. Applying Pythagorean theorem on triangle $A D P$, we have $x^{2}+(4 x)^{2}=16$, so that $x^{2}=16 / 17$. Hence, the area of square $A B C D$ is $16 x^{2}=256 / 17$.