Two equilateral triangles of side length $6$ centimeters are placed inside an $8$-by-$6$-centimeter rectangle. A side of each triangle coincides with one of the rectangle's opposite $6$-centimeter sides. What is the area of the triangles' intersection?

1. The intersection is a rhombus. Its horizontal diagonal runs between the triangle tips, so its length is $3\sqrt3-(8-3\sqrt3)=6\sqrt3-8$.
2. At the midpoint $x=4$ between the rectangle's opposite $6$-centimeter sides, the slanted sides meet at $y=3\pm(3-4/\sqrt3)$. Thus the vertical diagonal is $6-8/\sqrt3$.
3. The area is half the product of the diagonals:
   $$[\text{intersection}]=\frac12(6\sqrt3-8)(6-8/\sqrt3)=\frac{86\sqrt3}{3}-48$$
   square centimeters.