Thanks for sharing that formula, I've never come across it before.

The formula, however, does only work when the two medians (AD and BE in the formula) are perpendicular. Therefore, it doesn't apply to 6.4.

To understand why the formula works for AD and BE perpendicular, let's first rewrite the formula as $[ABC] = \dfrac{2}{3}\cdot AD\cdot BE$.

Using the usual notation, let $G$ be the intersection of $AD$ and $BE$ so $G$ is the centroid. First note that $[ABC] = 3\cdot [AGB]$ since the three medians divide the triangle into six triangles of equal area (and $\triangle AGB$ is made of two of these triangles).

To calculate $[AGB]$ we have$$[AGB] = \frac{1}{2}\cdot AG\cdot BG = \frac{1}{2}\cdot \frac{2}{3} AD\cdot \frac{2}{3} BE= \frac{2}{9} \cdot AD\cdot BE.$$Thus, $[ABC] = 3\cdot [AGB] = \dfrac{2}{3}\cdot AD\cdot BE$ as needed.

From this, it's not too hard to see that this ONLY works when $\triangle AGB$ is a right triangle, otherwise the area is not the same.