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24 Jul 2020, 04:49
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
35%
(medium)
Question Stats:
69% (01:50) correct
31%
(01:47)
wrong
based on 83
sessions
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A regular polygon will be drawn with a number of sides randomly chosen from the whole numbers 3 through 10. What is the probability that the internal angles will be greater than 110 ?
A) 1/8
B) 1/4
C) 3/8
D) 1/2
E) 5/8
A) 1/8
B) 1/4
C) 3/8
D) 1/2
E) 5/8
A regular polygon will be drawn with a number of sides randomly chosen from the whole numbers 3 through 10. What is the probability that the internal angles will be greater than 110 ?
A) 1/8
B) 1/4
C) 3/8
D) 1/2
E) 5/8
Angle at any vertice of the n sided polygon = \(\frac{180(n - 2)}{n}\)
As we know that upto a regular polygon of 5 side the angle is < 110.
From sides 6 to 10(total 5) have angle > 110
Angle at any vertice of the 6 sided polygon = \(\frac{180(6 - 2)}{6}\) = 120
Hence \(probability = \frac{required possibilities }{ total possibilities }\)
= 5/8
Two things to observe:
1. If you know how regular polygon works then you know that any regular polygon having 6 or more sides would have angle > 110. Visualise it.
2. If you can't visualise it, know from the formula \(\frac{180(n - 2)}{n}\) that numerator would increase in the multiples of 180 whereas denominator would increase by 1 only. Thus, if 6 sided regular polygon has angle > 110, others with more than 6 sides definitely would have angle more than 110.
Answer E.
A) 1/8
B) 1/4
C) 3/8
D) 1/2
E) 5/8
Angle at any vertice of the n sided polygon = \(\frac{180(n - 2)}{n}\)
As we know that upto a regular polygon of 5 side the angle is < 110.
From sides 6 to 10(total 5) have angle > 110
Angle at any vertice of the 6 sided polygon = \(\frac{180(6 - 2)}{6}\) = 120
Hence \(probability = \frac{required possibilities }{ total possibilities }\)
= 5/8
Two things to observe:
1. If you know how regular polygon works then you know that any regular polygon having 6 or more sides would have angle > 110. Visualise it.
2. If you can't visualise it, know from the formula \(\frac{180(n - 2)}{n}\) that numerator would increase in the multiples of 180 whereas denominator would increase by 1 only. Thus, if 6 sided regular polygon has angle > 110, others with more than 6 sides definitely would have angle more than 110.
Answer E.
24 Jul 2020, 06:01
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If you know a regular pentagon has 108 degree angles (not normally very useful to know), and a regular hexagon has 120 degree angles (often useful to know), then you can see we just need to have at least 6 sides. Since there are five values between 6 and 10 inclusive among the eight possible values between 3 and 10 inclusive, the answer is 5/8.
