How to find the sides of a polygon with 44 diagonals?
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heya mate here is your answer.
no. of diagonal for n sided polygon are
n(n-3)÷2
diagonals are 44
44=n(n-3)÷2
2 will get multiplied by 44
so
88=n(n-3)
solving n
n²-3n=88
n²-3n-88=0
factorising
(n-11)(n-8)=0
taking (n-8)
(n-8)=0
n=-8
can't use -8 because no. having - sign is not used
(n-11)=0
n=11
so the polygon have 11 sides
hope this helps you..
no. of diagonal for n sided polygon are
n(n-3)÷2
diagonals are 44
44=n(n-3)÷2
2 will get multiplied by 44
so
88=n(n-3)
solving n
n²-3n=88
n²-3n-88=0
factorising
(n-11)(n-8)=0
taking (n-8)
(n-8)=0
n=-8
can't use -8 because no. having - sign is not used
(n-11)=0
n=11
so the polygon have 11 sides
hope this helps you..
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