Question -
Answer -
(i) a hexagon
We know that a hexagonhas 6 angular points. By joining those any two angular points we get a linewhich is either a side or a diagonal.
So number of linesformed = 6C2
By using the formula,
nCr = n!/r!(n – r)!
6C2 = 6!/2!(6-2)!
= 6! / (2! 4!)
= [6×5×4!] / (2! 4!)
= [6×5] / (2×1)
= 3×5
= 15
We know number ofsides of hexagon is 6
So, number of diagonals= 15 – 6 = 9
The total no. ofdiagonals formed is 9.
(ii) a polygon of 16 sides
We know that a polygonof 16 sides has 16 angular points. By joining those any two angular points weget a line which is either a side or a diagonal.
So number of linesformed = 16C2
By using the formula,
nCr = n!/r!(n – r)!
16C2 = 16!/2!(16-2)!
= 16! / (2! 14!)
= [16×15×14!] / (2!14!)
= [16×15] / (2×1)
= 8×15
= 120
We know number ofsides of a polygon is 16
So, number ofdiagonals = 120 – 16 = 104
The total no. ofdiagonals formed is 104.