In the regular hexagon $A B C D E F$, two of the diagonals, $F C$ and $B D$, intersect at $G$. The ratio of the area of quadrilateral FEDG to the area of $\triangle B C G$ is
(A) $3 \sqrt{3}: 1$
(B) $4: 1$
(C) $6: 1$
(D) $2 \sqrt{3}: 1$
(E) $5: 1$

![](https://cdn.mathpix.com/cropped/2024_04_20_ac36362783317e0251fdg-139.jpg?height=214&width=258&top_left_y=1216&top_left_x=1389)
In the regular hexagon $A B C D E F$, two of the diagonals, $F C$ and $B D$, intersect at $G$. The ratio of the area of quadrilateral $F E D G$ to the area of $\triangle B C G$ is
(A) $3 \sqrt{3}: 1$
(B) $4: 1$
(C) $6: 1$
(D) $2 \sqrt{3}: 1$
(E) $5: 1$

![](https://cdn.mathpix.com/cropped/2024_04_20_33a8a6559d675f59baa2g-278.jpg?height=350&width=393&top_left_y=1294&top_left_x=1430)

## Solution 1

Join $E$ to $B$ and $D$ to $A$ as shown. Also join $E$ to $A$ and draw a line parallel to $A E$ through the point of intersection of $B E$ and $A D$. Quadrilateral $F E D G$ is now made up of five triangles each of which has the same area as $\triangle B C G$. The required ratio is $5: 1$.

![](https://cdn.mathpix.com/cropped/2024_04_20_33a8a6559d675f59baa2g-278.jpg?height=358&width=393&top_left_y=1691&top_left_x=1427)

## Solution 2

For convenience, assume that each side of the hexagon has a length of 2 units. Each angle in the hexagon equals $120^{\circ}$ so $\angle B C G=\frac{1}{2}\left(120^{\circ}\right)=60^{\circ}$. Now label $\triangle B C G$ as shown. Using the standard ratios for a $30^{\circ}-60^{\circ}-90^{\circ}$ triangle we have $B G=\sqrt{3}$ and $G C=1$.

![](https://cdn.mathpix.com/cropped/2024_04_20_33a8a6559d675f59baa2g-279.jpg?height=355&width=249&top_left_y=300&top_left_x=1491)

The area of $\triangle B C G=\frac{1}{2}(1) \sqrt{3}=\frac{\sqrt{3}}{2}$. Dividing the quadrilateral $F G D E$ as illustrated, it will have an area of $2(\sqrt{3})+\frac{1}{2}(1)(\sqrt{3})=\frac{5 \sqrt{3}}{2}$.

![](https://cdn.mathpix.com/cropped/2024_04_20_33a8a6559d675f59baa2g-279.jpg?height=263&width=371&top_left_y=698&top_left_x=1430)

The required ratio is $\frac{5 \sqrt{3}}{2}: \frac{\sqrt{3}}{2}$ or $5: 1$, as in solution 1 .

ANSWER: (E)