The $180^\circ$ rotation takes each line to a parallel line, so the two triangles overlap in a central hexagon

The centroid sits at $\frac{1}{3}$ of each height, so $A'$ lands at height $\frac{h}{3}$ beyond line $BC;$ each protruding corner is similar to the triangle with ratio $\frac{1}{3}$

A $180^\circ$ rotation takes each line to a parallel line, so $\triangle A'B'C'$ is congruent to $\triangle ABC$ with parallel sides. View $BC$ as horizontal and let $h$ be the height of $A$ above it. The centroid $G$ is at height $\frac{h}{3},$ so $A',$ the reflection of $A$ through $G,$ is at height $2 \cdot \frac{h}{3} - h = -\frac{h}{3},$ on the far side of line $BC,$ while $B'$ and $C'$ are at height $\frac{2h}{3}.$

Line $BC$ therefore slices off the corner of $\triangle A'B'C'$ at $A':$ the cut is parallel to $B'C',$ and the corner’s height $\frac{h}{3}$ is one third of the triangle’s full height $h,$ so the corner is similar with ratio $\frac{1}{3}$ and has area $\frac{1}{9}[ABC].$ The same happens at each side of $\triangle ABC,$ and these three corners are exactly the part of $\triangle A'B'C'$ outside $\triangle ABC.$ Hence the union has area $$\begin{aligned} &[ABC] + 3 \cdot \tfrac{1}{9}[ABC] \\ &= \tfrac{4}{3}[ABC]. \end{aligned}$$

By Heron’s formula with $s = 21,$ $[ABC] = \sqrt{21 \cdot 8 \cdot 7 \cdot 6} = 84,$ so the union has area $\frac{4}{3} \cdot 84 = 112.$