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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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Q: How many sides does a regular polygon have when it has 4752 diagonals showing work?

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A square or other rhombus (= parallelogram with 4 equal sides, whether or not they're at right angles).Also, any other regular 4n-gon -- i.e., regular polygon whose number of sides is a multiple of 4 -- will have pairs of perpendicular diagonals. So a regular octagon has two such pairs.

65

It works out that a polygon with 1175 diagonals has 50 sides

The formula is: 0.5*(n2-3n) where n is the number of sides of the polygon

If the sides and angles are equal, yes. Otherwise, no.

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