A polygon of n sides has n(n-3)/2 diagonals . Of a polynomial has 9 diagonals find the no of sides of the polygon
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There is polygon with exactly 9 diagonals has 6 sides
solution
no of diagonals=n(n-3)/2
given no of diagonals=9
by problem
n(n-3)/2=9
n(n-3)=18
nsquare -3n =18
nsquare -3n-18=0
nsquare +3n-6n-18=0
n(n+3)-6(n-3)=0
(n+3)(n-6)=0
n=-3 or 6
side can't be negative so it has 6sides
solution
no of diagonals=n(n-3)/2
given no of diagonals=9
by problem
n(n-3)/2=9
n(n-3)=18
nsquare -3n =18
nsquare -3n-18=0
nsquare +3n-6n-18=0
n(n+3)-6(n-3)=0
(n+3)(n-6)=0
n=-3 or 6
side can't be negative so it has 6sides
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