math probability! thank you!?

Question

the diagonals of a convex polygon are made by combining the vertices two at a time. However, some of the combimations are sides rather than diagonals. how many diagonals are there in a convex
a. pentagon (5sides)?
b. decagon(10 sides)?
c. n-gon(n sides)? simplify your answer as much as possible.

Answer 1

a. 5 (5*4/2 = 10 combinations - 5 that are sides = 5 diagonals)

b. 10*9/2 = 45 - 10 sides = 35 diagonals

c. n*(n-1)/2 - n = (n^2 -n)/2 -n = (n^2 -3n)/2

Check using pentagon (5^2-3*5)/2 = (25-15)/2 = 10/2 = 5 Checks - pentagon has five diagonals

Check using square (4^2-3*4)/2 = (16-12)/2 = 4/2 = 2
Checks - square has two diagonals

Check using triangle (3^2 - 3*3)/2 = (9-9)/2 = 0
Checks - triangles don't have diagonals

Answer 2

A) pentagon. There are 5 vertices, so there are 10 ways to connect them--the first vertex connects to all four of the others, the next to all three of the remaining ones, and so on. 4 + 3 + 2 + 1 = 10. Five of these are sides, so there are 5 diagonals.

B) decagon. There are 10 vertices, so there are 45 ways to connect them (9 + 8 + 7...). Ten of these are sides, so there are a total of 35 diagonals.

C) n-gon. There are n sides, so to find the number of ways to connect them, you need to sum the integers from 1 to n-1. That means there are (n-1)/2 pairs of numbers, each of which add up to n, so there are (n^2 - n)/2 ways to connect the vertices. Subtract n because there are n sides, and you get (n^2 - 3n)/2.