Why am I getting the wrong formula for the area of a dodecagon?

Question

More likely than not, I'm just making a simple algebraic mistake, but I can't seem to find it and so I would like some help.

Divide a (regular) dodecagon into $12$ congruent isosceles triangles with the length of the equal sides being the apothem $r$. The angle formed by two equal sides is equal to $30°$. Why, then, does the formula

$A = 12 \cdot r^2 \cos(15°) \sin(15°) =3r^2 \neq 12 \left(2-\sqrt{3} \right) r^2$ fail?

Answer 1

The length of the apothem (the length of a segment from the center of the dodecahedron to the midpoint of a side, or equivalently the radius of the inscribed circle) and the circumradius (the distance from the center to a vertex) are not equal. The area of a dodecahedron is $3R^2$, where $R$ is the circumradius, and $12(2-\sqrt3)r^2$ where $r$ is the apothem.

Answer 2

The apothem of a regular $n$-gon inscribed in a circle with radius $r$ is equal to $$r\cos\frac\pi n.$$

Here we have $$\cos\frac\pi{12}=\cos\Bigl(\frac\pi3-\frac\pi4\Bigr)=\frac{\sqrt 2(\sqrt 3+1)}4.$$