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 $\triangle B C G$ 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_e85ef69091c57cb633f3g-140.jpg?height=300&width=347&top_left_y=197&top_left_x=1344)
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 $\triangle B C G$ 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_688a3b42f69739f2e244g-259.jpg?height=350&width=395&top_left_y=286&top_left_x=1426)

## 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 $5: 1$.

## 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_688a3b42f69739f2e244g-259.jpg?height=360&width=247&top_left_y=1167&top_left_x=1492)

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 $2(\sqrt{3})+\frac{1}{2}(1)(\sqrt{3})=\frac{5 \sqrt{3}}{2}$.

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

![](https://cdn.mathpix.com/cropped/2024_04_20_688a3b42f69739f2e244g-259.jpg?height=258&width=357&top_left_y=1622&top_left_x=1445)

ANSWER: (E)