Re: The probability of drawing a regular polygon
[#permalink]
25 May 2020, 15:08
each internal angle of a regular polygon = (n-2)*180 / n {n: no of sides of polygon)
No of sides from 3 - 10, we have following values of the internal angle
n = 3; angle: (3-2)*180 / 3 = 60
n = 4; angle: (4-2)*180 / 4 =90
n = 5; angle: 3*180 / 5 = 108
n = 6; angle: 4*180 / 6 = 120, greater than 110
its clear that with n = 7, 8, 9 and 10; internal angle value is greater than 110
the probability that the internal angles will be greater than 110 = (no of polygon with angle > 110) / total polygons
no of the polygon with angle > 110: 5 {with sides 6-10}
the probability that the internal angles will be greater than 110 = 5/8
Ans: E