For each of the 8 vertices you can draw 5 diagonals and hence you have constructed \(5 \times 8 = 40\) diagonals.
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Let's examine ONE vertex in an octagon (an 8-sided polygon)

We can draw 5 diagonal lines from that one vertex


So, each of the 8 vertices can have 5 diagonal lines
So, the total number of diagonal lines = (8)(5) = 40

But wait!!

We have inadvertently counted every diagonal TWICE.
For example, we counted the diagonal from vertex A to vertex B AND we counted the diagonal from vertex B to vertex A

So, to account for this DUPLICATION, we must divide 40 by 2 to get 20

Answer: C

Cheers,
Brent
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#Formula for total number of diagonal of any regular polygon = n(n-3)/2 here n is number of sides
Number of diagonal can be drawn from any vertex = n-3

#length of diagonal of a square= s√2 here 's' is length of side

#length of Main diagonal of a cube= s√3
however, length of "face" diagonal of a cube = s√2


# length of diagonal of a rectangle = √(a^2+b^2) , here a & b are sides of rectangle
# length of diagonal of a Solid rectangle = √(a^2+b^2+c^2) ,here a,b & c are sides of solid rectangle

There are 8C2 combinations of lines between any two points in octagon. 8 combinations are the sides of octagon. Hence, 8C2-8=20. Answer is C