Question 568980
To calculate the area of regular polygon with N sides, the polygon can be sliced pizza-style into N isosceles triangles with one side as the base, and the area of those triangles can be calculated.
The height of those triangles is called the apothem, and can be calculated based on the length of the side, s, and half of the vertex angle, which depends on the number of sides, N.
{{{drawing(300,160,-2,28,-8,8,
triangle(0,7,0,-7,26.124,0),
line(0,0,26.124,0),
locate(8,0,apothem), locate(-1.2,1.1,s),locate(20,1.5,B),
arrow(-1,0.8,-1,7),arrow(-1,-0.8,-1,-7)
)}}} For a dodecagon, the half of the vertex angle is {{{B=360^o/24}}}
Putting it all together, the area, {{{A}}}, of a regular polygon with {{{N}}} sides of length {{{s}}} is
{{{A=s^2N/(4tan(pi/n))}}} or {{{A=s^2N/(4tan(180^o/n))}}} if you prefer your angles in degrees.
With {{{N=12}}} , {{{A=s^2*12/(4tan(180^o/12))=3s^2/tan(15^o)}}} --> {{{Atan(15^o)=3s^2}}} --> {{{s^2=Atan(15^o)/3}}}
{{{tan(15^o)=2-sqrt(3)}}} as can be calculated from {{{tan(45^o)}}} and {{{tan(30^o)}}} using the formula (trigonometric identity) for tangent of a difference.
So {{{s^2=Atan(15^o)/3}}} with {{{A=24+12sqrt(3)=12(2+sqrt(3))}}} and {{{tan(15^o)=2-sqrt(3)}}} gives us
{{{s^2=12(2+sqrt(3))(2-sqrt(3))/3=12(2^2-(sqrt(3))^2)/3=4(4-3)=4*1=4}}}
The length of a side of the polygon is {{{s=2}}}
and the perimeter is {{{12*2=highlight(24)}}}