Stop going in circles

There is a general formula for a regular $n$ -gon inscribed in a unit circle, where the expected area of a triangle so specified is

$\dfrac{3n*\cot(\frac{\pi}{n})}{2(n - 1)(n - 2)}$ .

Plugging in $n = 8$ yields an expected area of $\dfrac{2(1 + \sqrt{2})}{7}$ .

Thus $a*b*c*d = 2*1*2*7 = \boxed{28}$ .

I won't prove this formula but I will provide a method later that is more laborious. For now I just wanted to get something posted in case anyone wants to make a comment or query.

EDIT: O.k., now for some details .....

Labeling the vertices clockwise as $1$ through $8$ , (the choice of which vertex is labeled $1$ being arbitrary), we can form $\binom{8}{3} = 56$ triangles, with a great deal of repetition in sizes. We have essentially $5$ different triangles, similar to one triangle with vertices $(a,b,c)$ as follows:

(i) $8$ triangles similar to $(1,2,3)$ , which has area $\frac{\sqrt{2}(2 - \sqrt{2})}{4}$ ;

(ii) $16$ triangles similar to $(1,2,4)$ , which has area $\frac{1}{2}$ ;

(iii) $16$ triangles similar to $(1,2,5)$ , which has area $\frac{\sqrt{2}}{2}$ ;

(iv) $8$ triangles similar to $(1,3,5)$ , which has area $1$ , and

(v) $8$ triangles similar to $(1,3,6)$ , which has area $\frac{1 + \sqrt{2}}{2}$ .

We can then take the weighted average of these triangle areas to obtain the aforementioned solution.

I'll leave it as an exercise for the reader to verify the area values for triangles (i) through {v} found above, (i.e., I'm getting tired of typing. :) )