The distance between two opposite vertices of the dodecagon is 2. Find the area of the dodecagon.

Question

The distance between two opposite vertices of a regular dodecagon is 2. Find the area of the dodecagon.

I drew a diagram but I don't have any ideas. Any help? Thanks in advance!

Answer 1

The area of a regular dodecagon is just three times the square of its circumradius:

So a regular dodecagon with diameter $2$ has area $\color{red}{\large 3}$.

Here it is a non-animated but colorful dissection:

Answer 2

Divide the regular dodecagon into twelve triangles by drawing line segments from the vertices to the center. Each triangle has two sides of length $1$ and the included angle between these sides measures $30°$.

Let $O$ be the center of the dodecagon and $A, B, C$ be three consecutive vertices. Draw diagonal $AC$ which subtends a central angle of $60°$. Triangle $OAC$ is isosceles with its base angle equal to that subtended $60°$ central angle, thus equilateral so $AC=OA=OC=1$. At the same time $OB$ is the perpendicular bisector of $AC$ so the altitude of triangle $OAB$ to side $OB$ is $AC/2=1/2$. Then the area of the triangle is $(1/2)\times(1)\times(1/2)=1/4$. With twelve such triangles we get the full dodecagon having an area of $3$ square units.