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Consider a regular polygon with n sides (and n vertices).

Select any vertex. This can be done in n ways.

There is no line from that vertex to itself.

The lines from the vertex to the immediate neighbour on either side is a side of the polygon and so a diagonal.

The lines from that vertex to any one of the remaining n-3 vertices is a diagonal.

So, the nuber of ways of selecting the two vertices that deefine a diaginal seem to be n*(n-3). However, this process counts each diagonal twice - once from each end.

Therefore a regular polygon with n sides has n*(n-3)/2 diagonals.

Now n*(n-3)/2 = 4752

So n*(n-3) = 9504

that is n2 - 3n - 9504 = 0

using the quadratic equation, n = [3 + sqrt(9 + 4*9504)]/2 = 99 sides/vertices.

The negative square root in the quadratic formula gives a negative number of sides and that answer can be ignored.

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βˆ™ 2012-09-30 17:56:16
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Q: How many sides does a regular polygon have when it has 4752 diagonals showing work?
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