Suppose the polygon's vertices are denoted A1A2...An. We can fix one endpoint of the diagonal at A1 and count the number of ways to choose from the remaining vertices. There are n-3 vertices to choose from, because we cannot count A1, nor count A2 or An (since they are directly next to A1).
There are n vertices total, so we multiply to get n(n-3). However, we have to divide by 2 because we count each diagonal twice. The diagonal AiAj is the same as AjAi, and each is counted. Therefore, the number of diagonals in an n-gon is n(n-3)/2.
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It is: 0.5*(n2-3n) = diagonals whereas 'n' is the number of sides of the polygon
You can find the number of diagonals in a polygon using the formula n(n-3)/2, where n is the number of sides. Therefore an 11 sided polygon has 44 diagonals.
Use the formula of: 0.5*(n2-3n) whereby n is the number of sides of the polygon
The formula used to find the number of diagonals in a polygon is n(n-3)/2. 30 x 27/2 = 810/2 = 405 diagonals.
nope.aviImproved Answer:-33 because 1/2*(332-99) = 495