The interior angles of a polygon are in A.P. The smallest angle is 120 degree and the common difference is 5degree. Find the number of sides of the polygon.
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Answer:
Smallest angle=120degree
Common difference=5
A P is 120, 125, 130,……..
The sum of interior angles of a polygon= (n-2)180
Hence Sum of n terms of an A P = (n-2)180
n/2 {2.120+(n-1)5} = 180(n-2)
5n^2 -125n +720 = 0
n^2 -25n +144=0
n=9 or 16
hence number of sides can be 9 or 16
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SOLUTION
Suppose their are n sides of the polygon. The sum of the interior angles is given by
Suppose their are n sides of the polygon. The sum of the interior angles is given bysn= (2n-4)π/2 = (n-2)π
Since, interior angles forms an A.P. with
a= 120°
d= 5°
sn= n/2[ 2a+(n-1)d]
=) sn= n/2[2×120° +(n-1)5°]
=) (n-1)180° =n/2[240° +(n-1)5°] [from1]
=) (n-2)× 360° = n(5n+235°)
=) n^2 -25 +144 = 0
=) n= 9,16
when n= 16, then the last angle
an= a+(n-1)d
=) 120° +(16-1) 5°
=) 120° + 15×5°
=) 120°+ 75°
=) 195°, which is not possible.
Hence, required value of n is 9.
hope it helps ☺️
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