The formula for finding the variety of diagonals in a n-sided convex polygon is:


But just how is this formula derived? exactly how would I desire to begin deriving this formula?


For each vertex $n $(vertices) you have actually $(n-3)$ possible vertices to produce a diagonal line from.

Why $(n-3)$? because you can"t combine with the 2 neighboring vertices and you can"t incorporate with your existing vertex. Division by two due to the fact that you are counting twice. Therefore you have actually $frac(n-3)n2$ diagonals.

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We can solve this by using combination, to make diagonals we require to choose no. Of pairs of 2 vertices that deserve to be created from $n$ vertices i.e $$ dbinomn2$$ us will get the answer, however we must subtract $n$ from it since adjacent vertices cannot do a diagonal.$ herefore$ the last answer is $$dbinomn2-n= dfrac(n-1)n-2n2=dfracn(n-3)2.$$


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