Extend line $AG$ to a point $H$, so that $GH = 3$. Then $GBHC$ is a parallelogram with sides $4$ and $5$, and one of its diagonals of length $3$ (why?).  What else can you say about this parallelogram? Can you find its area? How does the area of this parallelogram relate to the area of the triangle?

* Recall that the centroid splits each median in a $2:1$ ratio, and the $6$ triangles made up by the medians all have the same area.