Why do you think each letter in a polygon's name can be used only once? ____________________________________________________ ______________/\ _____________/-- \ ____________/ -----\ ___________/______\ Its called triangle :)
To avoid confusion as to which vertex or side you are talking about.
because every point must have its own name to identify it from others.
You use the same letter only once when naming a polygon (or parts of a polygon) to avoid ambiguity. If you have two line segments both labelled A, then you won't know which one you mean when you talk about segment A, right?
I want to say 2, once at each corner
why do each letter in the polygons name can only be used once??? why do each letter in the polygons name can only be used once???
That's not necessarily true. An octagon is an eight-sided polygon with two 'o's
Because every point must have its own name to identify it from
Why do you think each letter in a polygon's name can be used only once? ____________________________________________________ ______________/\ _____________/-- \ ____________/ -----\ ___________/______\ Its called triangle :)
To avoid confusion as to which vertex or side you are talking about.
because every point must have its own name to identify it from others.
You use the same letter only once when naming a polygon (or parts of a polygon) to avoid ambiguity. If you have two line segments both labelled A, then you won't know which one you mean when you talk about segment A, right?
I want to say 2, once at each corner
I think it is used once because it is no fair that one letter get to be used twice and the another letters are used once
Because if you gave more than one side the same letter - it would get confusing !
A polygon can be used infinitely many times!
An n-sided polygon wil have n*(n-3)/2 diagonals.Consider joining each vertex to every other vertex. That gives potentially n-1 vertices. However, two of these will be sides of the polygon and so not diagonals. So each vertex gives rise to (n-3) diagonals. There are n vertices in the polygon and so that gives n*(n-3) diagonals. But, this method counts each diagonal twice: once from each end and so the correct answer is n*(n-3)/2.