## Task B-1.7.

The length of the side of an equilateral triangle $A B C$ is $a$. Point $E$ is located on side $\overline{A B}$ and is $\frac{1}{3} a$ away from vertex $B$. Triangle $A^{\prime} B^{\prime} C^{\prime}$ is the axis-symmetric image of triangle $A B C$ with respect to the line passing through point $E$ and perpendicular to side $\overline{A B}$. If the perimeter of the resulting figure (the union of triangles $A B C$ and $A^{\prime} B^{\prime} C^{\prime}$) is $12 \mathrm{~cm}$, what is its area?
## First Solution.

![](https://cdn.mathpix.com/cropped/2024_05_30_ab4e7c3a83d03c059773g-05.jpg?height=626&width=688&top_left_y=635&top_left_x=593)

The image clearly marks the vertices or concludes that the given axis of symmetry results in the pentagon $A A^{\prime} C^{\prime} D C$.

Triangle $B^{\prime} B D$ is an equilateral triangle with side length $\frac{2}{3} a$.

$P\left(B^{\prime} B D\right)=\frac{1}{4}\left(\frac{2}{3} a\right)^{2} \sqrt{3}=\frac{1}{4} \cdot \frac{4 a^{2} \sqrt{3}}{9}=\frac{a^{2} \sqrt{3}}{9}$.

$$
\begin{aligned}
P\left(A A^{\prime} C^{\prime} D C\right) & =P(A B C)+P\left(A^{\prime} B^{\prime} C^{\prime}\right)-P\left(B^{\prime} B D\right) \\
& =2 \cdot \frac{a^{2} \sqrt{3}}{4}-\frac{a^{2} \sqrt{3}}{9}=\frac{7 \sqrt{3}}{18} a^{2}
\end{aligned}
$$

The perimeter of the resulting pentagon is $3|A B|+3\left|B A^{\prime}\right|=3 a+3 \cdot \frac{a}{3}=4 a$,

so $a=3 \text{ cm}$.

The desired area is $P\left(A A^{\prime} C^{\prime} D C\right)=\frac{7 \cdot 9 \sqrt{3}}{18}=\frac{7 \sqrt{3}}{2} \text{ cm}^{2}$.