Solution:

Each of the $7$ triangles has $J$ as a vertex, and the centroid of a triangle $JVW$ lies on the segment from $J$ to the midpoint of $\overline{VW},$ two-thirds of the way out. So the centroid heptagon is the image of the heptagon $S$ formed by the midpoints of $\overline{AB}, \overline{BC}, \ldots, \overline{GH}$ under a dilation centered at $J$ with ratio $\frac{2}{3},$ and its area is $\frac{4}{9}[S].$

Place the rectangle with $A = (0, 6),$ $B = (8, 0),$ $C = (19, 0),$ $D = (27, 6),$ $E = (27, 17),$ $F = (19, 23),$ $G = (8, 23),$ $H = (0, 17),$ so $J = (0, \tfrac{23}{2}).$ The midpoints are $(4, 3),$ $(\tfrac{27}{2}, 0),$ $(23, 3),$ $(27, \tfrac{23}{2}),$ $(23, 20),$ $(\tfrac{27}{2}, 23),$ $(4, 20).$ The vertical segments at $x = 4,$ $x = \tfrac{27}{2},$ and $x = 23$ have lengths $17,$ $23,$ and $17,$ cutting $S$ into two trapezoids of height $\tfrac{19}{2}$ and a triangle of height $4:$ $$[S] = 2 \cdot \frac{17 + 23}{2} \cdot \frac{19}{2} + \frac{17 \cdot 4}{2} = 380 + 34 = 414.$$

The requested area is $\frac{4}{9} \cdot 414 = 184.$