[Geometry]
Problem: Let $T=\frac{9}{17}$. When $T$ is expressed as a reduced fraction, let $m$ and $n$ be the numerator and denominator, respectively. A square pyramid has base $A B C D$, the distance from vertex $P$ to the base is $n-m$, and $P A=P B=P C=P D=n$. Compute the area of square $A B C D$.

Solution: By the Pythagorean Theorem, half the diagonal of the square is $\sqrt{n^{2}-(n-m)^{2}}=\sqrt{2 m n-m^{2}}$. Thus the diagonal of the square is $2 \sqrt{2 m n-m^{2}}$, and the square's area is $4 m n-2 m^{2}$. With $T=\frac{9}{17}, m=9, n=17$, and the answer is 450 .

Final answer: 450