Find the area of the polygon whose vertices are the solutions in the complex plane of the equation $x^7+x^6+x^5+x^4+x^3+x^2+x+1=0$ If the area of the polygon whose vertices are the solutions in the complex plane of the equation $x^7+x^6+x^5+x^4+x^3+x^2+x+1=0$ can be expressed as $\frac{a\sqrt b+c}{d}$.Find $a+b+c+d$

The polynomial is $\frac{x^8-1}{x-1}$ has roots $\operatorname{cis}(2\pi k/8)$ for $k \in \{1, \ldots, 7\}$.
Thus the value is the area of the regular octagon minus the area of a triangle formed by two adjacent sides.
The area of an octagon (by splitting into triangles) with radius $1$ is $8 \cdot \frac{1}{2} \cdot \frac{\sqrt{2}}{2} = 2\sqrt{2}$.
I am stuck here.The answer for $a+b+c+d=10$


\begin{align}
p_1&=(\tfrac{\sqrt2}2,\tfrac{\sqrt2}2)
,\\
p_2&=(0,1)
,\\
p_3&=(-\tfrac{\sqrt2}2,\tfrac{\sqrt2}2)
,\\\ \dots
\end{align}  
\begin{align} 
S&=
(p_{1x}-p_{3x})^2+\tfrac32(p_{1x}-p_{3x})(p_{2y}-p_{1y})
=\frac{3\sqrt{2}+1}{2}
,\\
&=
\frac{3k\sqrt{2}+k}{2k}
,\\
a&=3k,\quad b=2,\quad c=k,\quad d=2k
,\\
a+b+c+d&=6k+2,
\quad k\in\mathbb{R}
.
\end{align}  
In order to get $a+b+c+d=10$,
we need to have $k=\tfrac43$.
