A3. Given is heptagon $A B C D E F G$, where all sides have length 2. Moreover, it is given that: $\angle E=120^{\circ}, \angle C=\angle G=90^{\circ}$ and $\angle A=\angle B=\angle D=\angle F$. What is the area of the heptagon?
A) $10+2 \sqrt{2}$
B) $8+3 \sqrt{3}$
C) 14
D) $10+2 \sqrt{6}$
E) $8+3 \sqrt{6}$

![](https://cdn.mathpix.com/cropped/2024_04_17_a91c336121b9423a3a1eg-1.jpg?height=297&width=402&top_left_y=1765&top_left_x=1515)
A3. B) $8+3 \sqrt{3}$ The heptagon can be divided into two squares and three equilateral triangles, all with side lengths of 2. We know that the area of such a square is 4. Using the Pythagorean theorem, we can calculate the height of, for example, triangle $A B M$ and we find $\sqrt{2^{2}-1}=\sqrt{3}$. The area of the triangle is then $\frac{1}{2} \cdot 2 \cdot \sqrt{3}=\sqrt{3}$. The total area of the heptagon is thus $2 \cdot 4+3 \cdot \sqrt{3}=8+3 \sqrt{3}$.

![](https://cdn.mathpix.com/cropped/2024_04_17_7effb8aeb7220f8b3452g-1.jpg?height=294&width=394&top_left_y=1658&top_left_x=1522)