[Geometry]
Problem: Let $T=67$. Rectangles $F A K E$ and $F U N K$ lie in the same plane. Given that $E F=T$, $A F=\frac{4 T}{3}$, and $U F=\frac{12}{5}$, compute the area of the intersection of the two rectangles.

Solution: Without loss of generality, let $A, U$, and $N$ lie on the same side of $\overline{F K}$. Applying the Pythagorean Theorem to triangle $A F K$, conclude that $F K=\frac{5 T}{3}$. Comparing the altitude to $\overline{F K}$ in triangle $A F K$ to $\overline{U F}$, note that the intersection of the two rectangles will be a triangle with area $\frac{2 T^{2}}{3}$ if $\frac{4 T}{5} \leq \frac{12}{5}$, or $T \leq 3$. Otherwise, the intersection will be a trapezoid. In this case, using similarity, the triangular regions of $F U N K$ that lie outside of FAKE each have one leg of length $\frac{12}{5}$ and the others of lengths $\frac{16}{5}$ and $\frac{9}{5}$, respectively. Thus their combined areas $\frac{1}{2} \cdot \frac{12}{5}\left(\frac{16}{5}+\frac{9}{5}\right)=6$, hence the area of the intersection is $\frac{5 T}{3} \cdot \frac{12}{5}-6=4 T-6$. With $T=67$, the answer is therefore $\mathbf{2 6 2}$.

Final answer: 262