length of side of a polygon having same area as a triangle Let P be a 12-sided regular polygon and T be an equilateral triangle with its incircle having radius $1$. I the area of P is the same as of area of T, then the length of side of P is
(A) $\sqrt{\sqrt{3} \cot15^\circ} \qquad$ (B) $ \sqrt{\sqrt{3} \tan15^\circ}\qquad$ (C) $ \sqrt{3\sqrt{2} \tan15^\circ}\qquad$ (D) $\sqrt{3\sqrt{2} \cot15^\circ} $
My try:
Since incenter of T is $1$, its height is $3$ and length of sides are $ 2\sqrt3 $. So area of T = $ 3\sqrt{3}$.
Area of $P=12*\delta.$ Where $\delta$ is the area of a triangle formed by any two adjacent vertices and the center of P. Let $x$ be length of sides of P, then $ \delta = \frac{x}{2}*h$. Here $\frac{\frac{x}{2}}{h} =\tan 15^ \circ \Rightarrow h = \frac{x\cot 15^\circ}{2} $. Therefore Area of P is $ 3x^2\cot15^\circ$
Equating areas of T and P, we have $$x^2\cot 15^\circ = \sqrt{3} \Rightarrow x^2 = \sqrt{3}\tan 15^\circ $$
So, my answer is $x = \sqrt{\sqrt{3} \tan15^\circ}$, option (B). But in my book, answer is given as (A). Can anyone suggest if my calculation of wrong somewhere?

Equilateral triangle area is correct. Equating it to a 12 sided regular polygon area,
$$ 12\cdot \dfrac{x^2}{4}\cot  15^{\circ}= 3 \sqrt{3}\rightarrow x= \sqrt{\sqrt3 \tan 15^{\circ}}$$
Your answer is correct.
