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An n-sided polygon wil have n*(n-3)/2 diagonals.

Consider joining each vertex to every other vertex. That gives potentially n-1 vertices. However, two of these will be sides of the polygon and so not diagonals. So each vertex gives rise to (n-3) diagonals. There are n vertices in the polygon and so that gives n*(n-3) diagonals. But, this method counts each diagonal twice: once from each end and so the correct answer is n*(n-3)/2.

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Q: How many diagonals will an -sided polygon have?
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