Thinking of a regular dodecagon inscribed in a circle, we can see that it is formed by 12 isosceles triangles whose sides are circle's radius, circle's radius and dodecagon's side; in each of these triangles the angle opposed to the dodecagon's side is equal to #360^@/12=30^@#; the area of each of these triangles is #("side"*"height)/2#, we only need to determine the height perpendicular to the dodecagon's side to resolve the problem.
In the mentioned isosceles triangle, whose base is the dodecagon's side and whose equal sides are the circle's radii, whose angle opposed to the base (#alpha#) is equal to #30^@#, there's only a line drawn from the vertex in which the circle's radii meet ( point C ) that intercepts perpendicularly the dodecagon's side: this line bisects the angle #alpha# as well as defines the triangle's height between the point C and the point in which the base is intercepted ( point M ), as well as divides the base in two equal parts (all because the two smaller triangles so formed are congruents).
Since the two smaller triangles mentioned are right ones we can determine the height of the isosceles triangle in this way:
#tan (alpha/2)="opposed cathetus"/"adjacent cathetus"# => #tan (30^@/2)= ("side"/2)/"height"# => #height = "side"/(2*tan 15^@)#
Then we have
#S_(dodecagon)= 12*S_(triangle)=12*(("side")("height"))/2=6*("side")("side")/(2*tan 15^@)# => #S_(dodecagon)= 3*("side")^2/(tan 15^@)#
