How many diagonals does a octagon have?
20 diagonals For any n-sided polygon, the formula for the number of diagonals is (n/2) x (n-3), so for an octagon it is (8/2) x (8-3) = 4 x 5 = 20.
No, I think 5 diagonals would be the maximum from a particular vertex. Let's label the vertices [A through H]. For this example, they go clockwise, starting with A and going around to H, so looking at A: B & H are adjacent and both of the segments AB and AH are actually sides of the octagon, not diagonals. The remaining diagonals, from A are: AC AD AE AF AG
This assumes you mean a regular octagon. There are 4 pairs where the diagonals each skip two vertices. Then there are 4 sets of three parallel diagonals where one is the main diagonal (skips three vertices) and two skip just one vertex. For each of these sets of three, you can choose three pairs of two. That makes 12 pairs all up. So there are 4 + 12 = 16 pairs.
There is insufficient information for us to answer this question. Please edit the question to include more context or relevant information. What formula of an octagon are you referring to: the sum of the interior angles, the frustum of a regular octagon, the area, the perimeter, the number of diagonals? The possibilities are almost endless!