In the hexagon of the following figure, the measure of all the internal angles is $\alpha$, which is why it is called equiangular.

![](https://cdn.mathpix.com/cropped/2024_05_01_d02c2755ad3373bde08ag-09.jpg?height=453&width=703&top_left_y=1661&top_left_x=756)

It is known that the lengths of the segments $A B, B C, C D$ and $D E$ are $|A B|=4,|B C|=5,|C D|=2,|D E|=3$.

a) Calculate the value of $\alpha$.

b) Calculate $|E F|$ and $|F A|$.

c) Calculate the area of the hexagon.
Solution

a) The sum of the measures of the internal angles of a hexagon is $4 \times 180^{\circ}$. Therefore, $6 \times \alpha=4 \times 180^{\circ}$, and thus, $\alpha=120^{\circ}$.

b) Extending the segments $A F, E D$ and $B C$, we obtain the following figure:

![](https://cdn.mathpix.com/cropped/2024_05_01_d02c2755ad3373bde08ag-35.jpg?height=645&width=726&top_left_y=508&top_left_x=742)

Since $\measuredangle F A B=\measuredangle A B C=120^{\circ}$, it follows that $\measuredangle Q A B=\measuredangle Q B A=60^{\circ}$. We conclude that triangle $Q A B$ is equilateral. Similarly, we conclude that triangles $R D C$ and $P F E$ are equilateral. Furthermore, triangle $P Q R$ is also equilateral because all its internal angles are equal. The length of the side of the equilateral triangle $P Q R$ is

$$
|Q B|+|B C|+|C R|=4+5+2=11
$$

Therefore, we also have that

$$
11=|R P|=2+3+|E P|
$$

which implies that $|E P|=6$. Thus, $|E F|=|P F|=|E P|=6$, while $|Q A|=|A B|=4$. Finally, for $|Q P|=|Q A|+|A F|+|F P|=$ $4+|A F|+6$ to be equal to 11, it is necessary that $|F A|=1$.

![](https://cdn.mathpix.com/cropped/2024_05_01_d02c2755ad3373bde08ag-35.jpg?height=647&width=743&top_left_y=1839&top_left_x=739)
c) The area of the hexagon can be calculated as the area of triangle $P Q R$ minus the sum of the areas of triangles $Q A B, R C D$ and $P F E$. Recall that the area of an equilateral triangle with side length $\ell$ is $\ell^{2} \sqrt{3} / 4$. Therefore, the area of the hexagon is

$$
\frac{11^{2} \sqrt{3}}{4}-\left(\frac{4^{2} \sqrt{3}}{4}+\frac{2^{2} \sqrt{3}}{4}+\frac{6^{2} \sqrt{3}}{4}\right)=\left(11^{2}-4^{2}-2^{2}-6^{2}\right) \frac{\sqrt{3}}{4}=65 \frac{\sqrt{3}}{4}
$$