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

\$\$frac(n-3)n2\$\$

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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