In a quadrilateral, a student can draw two diagonals. in a pentagon, a student can draw five diagonals. in a hexagon, a student can draw nine different diagonals. how many diagonals can be drawn in a regular polygon that has 17 sides?

Hello,

Let's assume a polygonal of n vertices.

Starting from of vertice we can reach n-1 vertices by 2 must be excluded.

So there are n*(n-1-2) possibilities but each  possibility is counted twice

Number of diagonals = n(n-3)/2

Examples:
n=4==>#=4*(4-3)/2=4*1/2=2
n=5==>#=5(5-3)/2=5*2/2=5
n=6==>#=6*(6-3)/2=6*3/2=9

n=17==>#=17*(17-3)/2=119.

17 is c. congruent means equal in size

The answer is $9.90 with rounding to the nearest tenth