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for any polygon with 'n' number of sides, the number of diagonals will be

n(n-3) / 2

if n is 50, no. of diagonals will be 50(50-3)/2 = 47*25 = 1175

how do you get it?

take any polygon, say septagon i.e. with 7 sides

if you take 1 vertice, remainig vertices will be 7-1 that is 6.

there will be 2 consecutive vertices for any vertice, so here, this one has 2 consecutive vertices(its always like that) so the remaining vertices is 6-2 that is 4.

remember : a diagonal connects non consecutive vertices

so you know that from one vertice, there will be 4 diagonals.

multiply it by the total no. of vertices(here 7)

4*7 = 28

now divide this by 2

(since 2 vertices share the same diagonal)

therefore, 28/2 = 14 which is the no. of diagonals in a septagon.

so for an n-gon, the number of diagonals will be

n(n-3) / 2

hope it helps!

p.s. i figured it out myself during math class. then i tried it on many polygons.

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โˆ™ 2011-11-15 15:56:23
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