Question

(a) A polygon with n sides has a total of 1/p*n(n-q) diagonals, where p and q are integers.
(i) find thhe value of p and q
(ii) find the total number of diagonals in a polygon with 100 sides
(iii) find the number of sides of a polygon which has a total of 170 diagonals
(b) a polygon with n+1 sides has 30 more diagonals than a polygon with n sides. find n

Answer 1

100 sides polygon have 50 diagonals

Answer 2

(i) I'm still working on this
(ii) n=100 => [n(n-3)]/2 = [100(100-3)]/2 = 4850
(iii) [n(n-3)]/2 =170 => n2-3n-340=0 <=> n = 20 and n=-17. But sides couldn't be negative => n=20 (sides)
b) Call the total number of diagonals of n+1 sides polygon is x
=> [n(n-3)]/2=x-30 and [(n+1)(n+1-3)]/2=x
<=> n=31

Answer 3

I) p= 2
q= 3

Answer 4

Find the total number of diagonals in a polygon with 100 sides

Formula to find the number of diagonals

D= n(n-3) / 2. Where n is the number of sides

D = 100(100-3)/2
D= 100 x 97 / 2
D = 9700/2
D= 4850