Dodecagon Area Question The distance between two opposite vertices of the dodecagon is 2. Find the area of the dodecagon.
Is there any way to do this without trigonometry?
And could you include a proof also?
:O
 A: 
Obviously we have to assume it is a regular dodecagon, otherwise its area is indeterminate. I follow your idea of taking a circle through the vertices. We have $\angle A_1OA_2 = 30^o$, so $\angle A_1OA_3 = 60^o$. Also $OA_1 = OA_3$, so the triangle $A_1OA_3$ is equilateral. We are given that $OA_1=1$, so $A_1A_3 = 1$. $OA_2$ intersects $A_1A_3$ at $M$ the midpoint of $A_1A_3$. So $A_1M = \frac12$.
The area of triangle $A_1OA_2$ is half its base $OA_2$ times its height $A_1M$ and hence $\frac14$. There are 12 triangles, so the total area of the dodecagon is 3. 
A: This animated gif gives a proof without words that the area of a regular dodecagon is three times the square of its circumradius:

A: Another form of the area for a regular dodecagon that doesn't explicitly contain trigonometric functions, is $A = 3R^2$ where $A$ is area and $R$ is the circumradius (distance from center to vertex).
Therefore you have that the circumradius = $1$ (half the distance between opposite vertices) hence the area $A=3\cdot1^2=3$.
You can prove this by a cute tiling argument:
http://demonstrations.wolfram.com/KurschaksDodecagon/
