## Task B-1.7.

From point $D$ on side $\overline{A B}$, where $|A D|>|B D|$, of an equilateral triangle $A B C$, a perpendicular is drawn to side $\overline{B C}$ with its foot at $E$. Then, from $E$, a perpendicular is drawn to $\overline{C A}$ with its foot at $F$, and from $F$, a perpendicular is drawn to $\overline{A B}$ with its foot at $G$. The area of quadrilateral $D E F G$ is $21 \sqrt{3} \mathrm{~cm}^{2}$, and the length of the side of triangle $A B C$ is $14 \mathrm{~cm}$. Determine the length $x=|B D|$ and the perimeter of quadrilateral $D E F G$.
## Solution.

![](https://cdn.mathpix.com/cropped/2024_05_30_63d9d2269da4cabd8f72g-05.jpg?height=594&width=671&top_left_y=1285&top_left_x=590)

Each of the triangles $D B E, E C F$, and $F A G$ is half of an equilateral triangle.

Therefore, we have

$$
\begin{gathered}
|B E|=\frac{1}{2}|B D|=\frac{x}{2} \quad \text { so } \quad|C E|=14-\frac{x}{2} \\
|C F|=\frac{1}{2}|C E|=7-\frac{x}{4} \\
|F A|=14-|C F|=14-\left(7-\frac{x}{4}\right)=7+\frac{x}{4}
\end{gathered}
$$

Express the area of quadrilateral $D E F G$ in terms of $x$:

$$
\begin{aligned}
P_{D E F G} & =P_{A B C}-P_{D B E}-P_{E C F}-P_{F A G} \\
& =\frac{|A B|^{2} \sqrt{3}}{4}-\frac{1}{2} \cdot \frac{|D B|^{2} \sqrt{3}}{4}-\frac{1}{2} \frac{|C E|^{2} \sqrt{3}}{4}-\frac{1}{2} \cdot \frac{|F A|^{2} \sqrt{3}}{4} \\
& =\frac{14^{2} \sqrt{3}}{4}-\frac{1}{2} \cdot \frac{x^{2} \sqrt{3}}{4}-\frac{1}{2} \frac{\left(14-\frac{x}{2}\right)^{2} \sqrt{3}}{4}-\frac{1}{2} \cdot \frac{\left(7+\frac{x}{4}\right)^{2} \sqrt{3}}{4}
\end{aligned}
$$

Thus, we have

$$
\begin{gathered}
21 \sqrt{3}=\frac{\sqrt{3}}{8}\left(2 \cdot 14^{2}-x^{2}-\left(14-\frac{x}{2}\right)^{2}-\left(7+\frac{x}{4}\right)^{2}\right) \\
21 \cdot 8=2 \cdot 196-x^{2}-14^{2}+14 x-\frac{x^{2}}{4}-49-\frac{7 x}{2}-\frac{x^{2}}{16} \\
0=21+\frac{21}{2} x-\frac{21}{16} x^{2}
\end{gathered}
$$

We obtained the quadratic equation $x^{2}-8 x+16=0$

whose only solution is $x=4$, so $|B D|=4 \mathrm{~cm}$.

Further, we have

$$
\begin{aligned}
& |D E|=\frac{x \sqrt{3}}{2}=2 \sqrt{3} \mathrm{~cm} \\
& |E F|=\frac{\left(14-\frac{x}{2}\right) \sqrt{3}}{2}=6 \sqrt{3} \mathrm{~cm} \\
& |F G|=\frac{\left(7+\frac{x}{4}\right) \sqrt{3}}{2}=4 \sqrt{3} \\
& |G D|=|A B|-|D B|-\frac{1}{2}|F A|=14-4-\frac{1}{2} \cdot 8=6 \mathrm{~cm}
\end{aligned}
$$

The perimeter of quadrilateral $D E F G$ is $6+12 \sqrt{3} \mathrm{~cm}$.

## SCHOOL/CITY COMPETITION IN MATHEMATICS

2nd grade - high school - B variant
January 3, 2018.

## IF A STUDENT HAS A DIFFERENT APPROACH TO SOLVING THE PROBLEM, THE COMMITTEE IS OBLIGED TO GRADE AND EVALUATE THAT APPROACH APPROPRIATELY.