a polygon of n sides has n(n-3)by2 diagonals .if a polygon has 9 diagonals. find the number of sides of the polygon
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n (n-3)/2=9
n (n-3)=9*2
n (n-3)=18
n^2 -3n =18
n^2 -3n -18=0
(n-6)(n+3)=0
n=6& n= -3
as side of a polygon can't be negative
hence number of sides of the polygon is 6.
n (n-3)=9*2
n (n-3)=18
n^2 -3n =18
n^2 -3n -18=0
(n-6)(n+3)=0
n=6& n= -3
as side of a polygon can't be negative
hence number of sides of the polygon is 6.
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