[Geometry]
Problem: Let $T=20$. The lengths of the sides of a rectangle are the zeroes of the polynomial $x^{2}-3 T x+T^{2}$. Compute the length of the rectangle's diagonal.

Solution: Let $r$ and $s$ denote the zeros of the polynomial $x^{2}-3 T x+T^{2}$. The rectangle's diagonal has length $\sqrt{r^{2}+s^{2}}=\sqrt{(r+s)^{2}-2 r s}$. Recall that for a quadratic polynomial $a x^{2}+b x+c$, the sum of its zeros is $-b / a$, and the product of its zeros is $c / a$. In this particular instance, $r+s=3 T$ and $r s=T^{2}$. Thus the length of the rectangle's diagonal is $\sqrt{9 T^{2}-2 T^{2}}=T \cdot \sqrt{7}$. With $T=20$, the rectangle's diagonal is $\mathbf{2 0} \sqrt{\mathbf{7}}$.

Final answer: $20 \sqrt{7}$