Question

a polygon has 27 diagonals how many sides does it have

Answer 1

no. of diagonals=27

n(n-3)/2=27

n(n-3)=54

n2-3n=54

n2-3n-54=0

n2-9n+6n-54=0

n(n-9)+6(n-9)=0

(n+6)(n-9)=0

n+6=0 n-9=0

n=-6 n=9

hence, no. of diagonals cannot be in negative

so, n=9

therefore, no of sides for 27 diagonals is 9

Answer 2

no.of diagonals = 27

= n(n-3) = 54

= sqN -3n = 54

= sqN - 3n - 54 = 0

= sqN - 9n - 6n - 54 = 0

= n (n-9)+(n-9)=0

= (n-9)+6(n+6) = 0

= (n-9) (n+6) = 0

i) n-9 = 0

therefore = 9

ii) n+6 = 0

therefore = -6

( The result is always positive )

= 9 ,

( Ans )