I can then calculate the perimeter by multiplying the side length, x, but the number of sides, n.
I will use a hexagon as an example:
x is the side length of the hexagon. It is also the base of the triangles the hexagon is made up of:
As a hexagon has six sides, I can work out the angles that are at the bottom by taking 2 away from the number of sides, so 4, and multiplying this by 90:
a = 90 (n - 2)
Using the bottom angles, I can use trigonometry to calculate the height of the triangle. I do this by splitting the triangle into two right angled triangles:
This is done using this formula:
h = tan60 * (x/2)
If I want to use this formula so that it will work for any regular polygon, I can put the formula for an angle into it:
h = tan[(180(n-2))/2n] * (x/2)
Using the height, I can work out the area of the individual triangle:
a ∆ = (h * x)/2
Which, with the formula for height added, looks like this:
a ∆ = tan[(180(n-2))/2n] * (x/2) * (x/2)
To find the area of the total shape, I will need to multiply the above formula by the number of triangles there are, which is also the number of sides, n:
a = tan[(180(n-2))/2n] * (x/2) * (x/2) * n
The perimeter of the total shape can be worked out by multiplying the length of the sides by the number of them:
p = nx
To work out the side length of an equable shape, therefore, I need to put the two formulas together and rearange it so I am able to calculate x:
nx = tan[(180(n-2))/2n] * (x/2) * n * (x/2)
nx = tan[(90(n-2))/n] * (x2/4) * n
x = tan[(90(n-2))/n] * (x2/4)
4x = x2tan[(90(n-2))/n]
4 = x tan[(90(n-2))/n]
x = 4/tan[(90(n-2))/n]
This formula allows me to calculate the side length of any sided regular polygon.
Examples
Triangle
x = 4/tan[(90(n-2))/n]
x = 4/tan[(90(3-2))/3]
x = 4/tan[90/3]
x = 4/tan30
x = 4/0.58…
x = 6.90cm (2dp)
Square
x = 4/tan[(90(n-2))/n]
x = 4/tan[(90(4-2))/4]
x = 4/tan[180/4]
x = 4/tan45
x = 4/1
x = 4cm
Octagon
x = 4/tan[(90(n-2))/n]
x = 4/tan[(90(8-2))/8]
x = 4/tan[540/8]
x = 4/tan67.5
x = 4/2.31…
x = 1.66cm (2dp)
673 Sided shape
x = 4/tan[(90(n-2))/n]
x = 4/tan[(90(673-2))/673]
x = 4/tan[60390/673]
x = 4/tan89.73…
x = 4/214.22…
x = 0.019cm (2sf)
To conclude, I have discovered a formula that will let me discover the side length required for an equible shape of any amount of sides. I think that my formulas will be reliable for any amount of sides, this is demonstrated by my example of a random large number of sides. However, there is a problem it will give an answer for shapes with 1 side, saying the length should be -7, which is impossible.