Why does (n(n-3))/2 always gives the number of diagonals in a polygon with n sides?
Goldwaagefhu
Answered question
2023-01-02
Answer & Explanation
Digngoorpr43
Beginner2023-01-03Added 14 answers
In an n-sided polygon, you have n starting points for diagonals. And each diagonal can go to (n – 3) ending points because a diagonal can’t end at its own starting point or at either of the two neighboring points.
Now, the first step is to multiply n by (n – 3).
After, as each diagonal's closing point can be employed as a beginning point too the product n(n – 3) counts each diagonal twice. That’s why you divide by 2.
So number of diagonals becomes n(n-3)/2
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