Let $M$ be the midpoint of $A C$. For any triangle, we know that the centroid is located $2 / 3$ of the way from the vertex, so we have $M G / M B=M G^{\prime} / M B^{\prime}=1 / 3$, and it follows that $M G G^{\prime} \sim M B B^{\prime}$. Thus, $G G^{\prime}=B B^{\prime} / 3$. However, note that $B B^{\prime}$ is twice the altitude to $A C$ in triangle $A B C$. To finish, we calculate the area of $A B C$ in two different ways. By Heron's Formula, we have $$[A B C]=\sqrt{18(18-9)(18-10)(18-17)}=36$$ and we also have $$[A B C]=\frac{1}{4} B B^{\prime} \cdot A C=\frac{17}{4}(B B^{\prime})$$ from which it follows that $G G^{\prime}=B B^{\prime} / 3=48 / 17$.