In the diagram, $\triangle A B C$ is equilateral, $B C=2 C D, A F=6$, and $D E F$ is perpendicular to $A B$. What is the area of quadrilateral $F B C E$ ?
(A) $144 \sqrt{3}$
(B) $138 \sqrt{3}$
(C) $126 \sqrt{3}$
(D) $108 \sqrt{3}$
(E) $66 \sqrt{3}$

![](https://cdn.mathpix.com/cropped/2024_04_20_e85ef69091c57cb633f3g-155.jpg?height=342&width=482&top_left_y=577&top_left_x=1407)
Solution

Drop a perpendicular from $A$ to $B C$, and label as shown. Since $\mathrm{D} A B C$ is equilateral, $B N=N C=C D$. Let $B N=x$ and $B F=y$.

Then $6+y=2 x$.

Also, $-F A M=30 \mathrm{r}$, and $\mathrm{DAMF}$ is a $30 \mathrm{r}-60 \mathrm{r}-90 \mathrm{r}$ triangle with sides in the ratio $1: \sqrt{3}: 2$.

![](https://cdn.mathpix.com/cropped/2024_04_20_688a3b42f69739f2e244g-292.jpg?height=342&width=477&top_left_y=1325&top_left_x=1431)

Thus, $A M=4 \sqrt{3}$ and $F M=2 \sqrt{3}$.

Use similar triangles $D B F$ and $A M F$ to find

$$
\begin{aligned}
\frac{D B}{B F} & =\frac{A M}{M F} \\
\frac{3 x}{y} & =\frac{4 \sqrt{3}}{2 \sqrt{3}} \\
3 x & =2 y
\end{aligned}
$$

Solving equations (1) and (2) yields $x=12$ and $y=18$.

Find $A N=12 \sqrt{3}$; the area of $D A B C$ is thus $144 \sqrt{3}$.

Use similar triangles $E A F$ and $A M F$ to find $F E=6 \sqrt{3}$; the area of $D A F E$ is $18 \sqrt{3}$.

The area of quadrilateral $F B C E$ is $126 \sqrt{3}$.

ANSWER: (C)

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