Skip over navigation
Hexagon Perimeter
Age 14 to 16
ShortChallenge Level 
Drawing lines from the centre to each vertex of the hexagon, as shown, splits it into 6 identical isosceles triangles.
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.
$\tan {30^\text o}=\dfrac{\frac{s}{2}}{1}\rightarrow\tan{30^\text o}=\frac{s}{2}\rightarrow2\tan{30^\text o}=s$.
$\tan{30^\text o}=\dfrac{1}{\sqrt3}$, 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 $s$.
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 $6\times1.155=6.93$.
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.
$\tan {30^\text o}=\dfrac{\frac{s}{2}}{1}\rightarrow\tan{30^\text o}=\frac{s}{2}\rightarrow2\tan{30^\text o}=s$.
$\tan{30^\text o}=\dfrac{1}{\sqrt3}$, 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 $s$.
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 $6\times1.155=6.93$.
You can find more short problems, arranged by curriculum topic, in our short problems collection.
You may also like
Some(?) of the Parts
A circle touches the lines OA, OB and AB where OA and OB are perpendicular. Show that the diameter of the circle is equal to the perimeter of the triangle
Ladder and Cube
A 1 metre cube has one face on the ground and one face against a wall. A 4 metre ladder leans against the wall and just touches the cube. How high is the top of the ladder above the ground?
Bendy Quad
Four rods are hinged at their ends to form a convex quadrilateral. Investigate the different shapes that the quadrilateral can take. Be patient this problem may be slow to load.