Please do not thing of the figure above as a decagon, but as a polygon with *n* sides — an *n*-gon, in other words. How many diagonals does it have? Well, first, there are* n* vertices for diagonals to come from, and three vertices they cannot go to — themselves, and the two immediately on either side of them, since the segments to those vertices are sides, not diagonals. That’s* n* vertices firing diagonals at *n*-3 other vertices, or (*n*)(*n-*3). However, that counts each diagonal exactly twice (once from each side), so the actual number of diagonals is half that: ** d=(n)(n-3)/2**.

Now we can look at the polygon above, and use it to check this formula, by “remembering” that it is a decagon. With* n* = 10, *d* = (*n*)(*n*-3)/2 = (10)(7)/2 = 35.

Are there really 35 diagonals in the decagon above? Well, I made those of the same length into color-groups, to make them easier to count. There are five green ones, ten yellow ones, ten red ones, and ten pink ones, which does indeed total 35.

Suppose you know a polygon has, say, 104 diagonals. Can this formula be used to find the number of sides? Yes! Substituting 104 for *d* leads to this: 104 = (*n*)(*n*-3)/2, which then becomes 208 = (*n*)(*n*-3) = *n²* – 3*n*. To set this up for the quadratic formula, I’m rearranging it to* n² *– 3*n* – 208 = 0. The quadratic formula then states that* n* = (3 ± sqrt(9 – (4)(1)(-208)))/2 = (3 ± sqrt(9 + 832))/2 = (3 ± sqrt(841))/2 = (3 ± 29)/2 = (32 or -26)/2 = 16 or -13, and only one of these answers, 16, can be the number of sides of a polygon. Voila!

Such an easy method …. its fab

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wow

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it works

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