Triangles/Diagonals question

[yo4561]

On page 455 in all the quant, the book discusses how a hexagon can be cut into four triangles and shows you how to do it. Will this always be the case that you can only draw four triangles based on the number of vertices of a hexagon?

My second question relates to figuring out how many diagonals there are in a polygon. Is there a general formula we can use?

Thank you

[esledge]

I don't have that book in front of me, but I think the 4 triangles in the hexagon are created with dotted lines that zig-zag back and forth in the hexagon?

That's one way, but I think it's helpful to always split a shape into triangles the same way: pick one corner, and draw all of your dotted lines from it to the opposite corners. Do this with a 4-sided, a 5-sided, a 6-sided, and a 7-sided figure, and you'll see the pattern. You can't draw a dotted line to the corners adjacent to the corner you start from (or to the starting corner itself), so in a hexagon (with 6 sides), you draw 3 dotted lines (or cut-lines), thereby creating 4 triangles. Do this with the other figures I suggested, and you'll see the proof that you always get (n-2) triangles from an n-sided polygon, which is the proof for the rule that Sum of Interior Angles of a Polygon = (n-2)*180.

Your second question about "how many diagonals" is possibly different, if you are considering every possible diagonal (even those that criss-cross one another). In my 15 years teaching GMAT, I've seen one such problem, and I don't remember now whether it was even an official question. That's more of a Combinatorics question than a Geometry question, so let me know if that's what you meant and if you still want to talk about it.

But if you just meant "how many diagonals" are necessary to split a polygon into triangles with the fewest "cuts," as we discussed above, the answer is n-3 diagonals in an n-sided figure. (See the figures you drew earlier for proof!)

[yo4561]

Thank you Emily! To clarify, I came across this formula for the number of diagonals in a polygon but noticed you said (n-3).

The formula I found was (n(n-3)/2). Is this formula incorrect then?

[esledge]

Based on that n(n-3)/2 formula, that's every diagonal, even those that crisscross each other. Thanks for clarifying!

Yes, the formula is correct. Here's why:

Pick a vertex of a polygon. You can draw n-3 diagonals from it; the minus 3 is due to not drawing to that 1 corner itself nor drawing to the 2 adjacent vertices (those are sides of the polygon already).

Repeat the drawing exercise from all n vertices, so that's now n*(n-3) lines sketched and counted. However, this number double counts each diagonal: you counted a diagonal when you went from point A to point B, but also again when you drew from point B to point A. Thus, the division by 2.

This formula assumes that the polygon is convex, though. In a concave polygon, this formula would count some fake "diagonals" that lie outside the polygon partially or altogether.