Because $CD = AF - AC - DF = 14$, the side lengths of $\triangle CDG$ are $14$, $30$, and $40$. By Heron's Formula,
$$\text{Area}(\triangle CDG) = \sqrt{42(42 - 14)(42 - 30)(42 - 40)} = 168,$$
implying that the distance from $G$ to line $CD$ is $\frac{2\cdot168}{CD} = 24$. Then because $BE = BD + CE - CD = 39$, the area of $\triangle BGE$ is $\frac{1}{2} \cdot 39 \cdot 24 = 468$.

OR
As in the first solution, $CD = 14$. Let $H$ be the projection of $G$ onto line $AF$. Note that
$$CD^2 + DG^2 = 14^2 + 30^2 = 1096 < 1600 = CG^2,$$
implying that $\angle CDG$ is obtuse, and it follows that $D$ lies between $C$ and $H$. Applying the Pythagorean Theorem to $\triangle CHG$ and $\triangle DHG$ yields
$$40^2 = CH^2 + GH^2 = (DH + 14)^2 + GH^2 \quad \text{and}$$
$$30^2 = DH^2 + GH^2.$$
Subtracting the second equation from the first and dividing by $28$ gives $25 = DH + 7$, so $DH = 18$ and $GH = 24$. The result then follows as in the first solution.