Hexagon perimeter

Drawing lines from the centre to each vertex of the hexagon, as shown, splits it into 6 identical isosceles triangles.

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The total angle at the centre is $360^\text o$, so the angle at the centre of each triangle must be $360\div6 = 60^\text o$.

Using trigonometry

Since the angle at the middle of each triangle is $60^\text o$, when the triangle is split in half by the radius, as shown below, the angle will be $30^\text o$. In the diagram, $s$ denotes the side lengthof the hexagon.

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$\tan {30^\text o}=\dfrac{\frac{s}{2}}{1}\rightarrow\tan{30^\text o}=\frac{s}{2}\rightarrow2\tan{30^\text o}=s$.

$\tan{30^\text o}=\dfrac{1}{\sqrt3}$, so $s=2\times\dfrac{1}{\sqrt3}=\dfrac{2}{\sqrt{3}}$ or $1.155$.

The hexagon has $6$ sides, so its perimeter is $6\times\dfrac{2}{\sqrt3}=2\times\sqrt3\times\sqrt3\times\dfrac{2}{\sqrt3}=4\sqrt3$, or $6\times1.155=6.93$.

Using Pythagoras' Theorem

Since the angle at the middle of each triangle is $60^\text o$, the triangles must be equilateral triangles (as they are already isosceles, so the other two angles are equal, and add up to $120^\text o$).

The diagram below shows one of the triangles cut in half by the radius, where the side length of the hexagon is labelled $s$.

 

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Applying Pythagoras' theorem, $$\begin{align}&1^2+\left(\frac{s}{2}\right)^2=s^2\\

&\Rightarrow1+\dfrac{s^2}{4}=s^2\\

&\Rightarrow4+s^2=4s^2\\

&\Rightarrow4=3s^2\\

&\Rightarrow s^2=\frac{4}{3}\\

&\begin{split}\Rightarrow s&=\sqrt{\frac{4}{3}}\\

&=\dfrac{2}{\sqrt3}\approx1.155\end{split}\end{align}$$

The hexagon has $6$ sides, so its perimeter is $6\times\dfrac{2}{\sqrt3}=2\times\sqrt3\times\sqrt3\times\dfrac{2}{\sqrt3}=4\sqrt3$, or $6\times1.155=6.93$.