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 $a+b+c+d=10$

Let us use Roots of unity!
Here's a hint (by the way this question is from the ARML and I recommend you try it out because it is quite nice). We know that if $z\neq1$ is an $z$th root of unity, that $1+z+z^2+z^3+z^4+\cdots+z^{n-1}=0$. Also, note that the $n$th roots of unity form a regular $n$-gon with a vertex at $(1,0)$.
ANSWER (SPOILER):
After all this you will get an answer of $\frac{6\sqrt{2}+1}{4}$, so that $a+b+c+d=13$, so our answer is $\boxed{13}$
