One more math question?

Question

How many diagonals does a 12 sided figure have? (Like a triangle has none, pentagon has five diagonals etc. etc.) PLEASE EXPLAIN thank you very much for your time and effort (if you put any in)

Answer 1

Assuming a convex figure (because it's not true that an arbitrary pentagon has 5 diagonals unless you count "diagonals" that travel outside the pentagon itself), the diagonals of the figure are precisely those line segments that travel from one vertex to another vertex not adjacent to it. For an n-sided figure, there are n vertices, and so n choices for where a diagonal can start, and given a starting vertex, there are n-3 vertices that it can end on (n-3, since to be a diagonal the segment can't end on either the same vertex or either of the two adjacent ones). Therefore, if we count the diagonals from P to Q as distinct from the diagonal from Q to P, then there would be n*(n-3) diagonals of a convex n-gon. However, since the direction actually doesn't matter (i.e. the diagonal from P to Q and the diagonal from Q to P are the same), each diagonal is actually counted twice with that scheme, so in fact there are only n*(n-3)/2 diagonals in a convex n-gon.

Applying this formula to a convex dodecagon, we see that there are 12*(9)/2 = 54 diagonals.

Answer 2

starting from any given corner, there will be 3 less diagonals than there are sides, the next corner will have the same number. From there, as you continue around the shape, the number of new diagonals will decrease by one.