27. Let $A B C D E F$ be a regular hexagon. Let $G$ be a point on $E D$ such that $E G=3 G D$. If the area of $A G E F$ is 100 , find the area of the hexagon $A B C D E F$.
27. Answer: 240.
We may assume that $D E=1$ and denote $\lambda=E G=34$.
$$
\frac{E G A X}{D E X Y}=\frac{1+\lambda}{4} \text { implies } \frac{A G E F}{D E X Y}=\frac{1+\lambda}{4}-\frac{1}{8}=\frac{1+2 \lambda}{8} \text {. }
$$

Note that $\frac{A B C D E F}{D E X Y}=\frac{6}{8}$. Then $\frac{A G E F}{A B C D E F}=\frac{1+2 \lambda}{6}=\frac{5}{12}$.
If $A G E F=100$, then $A B C D E F=100 \times \frac{12}{5}=240$.