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What is the number of sides of a regular polygon in which 1/3rd of the

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What is the number of sides of a regular polygon in which 1/3rd of the  [#permalink]

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New post 23 Aug 2018, 00:42
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What is the number of sides of a regular polygon in which 1/3rd of the sum of exterior angle is equal to the each interior angle ?

(A) 3

(B) 4

(C) 5

(D) 6

(E) 8

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Re: What is the number of sides of a regular polygon in which 1/3rd of the  [#permalink]

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New post 23 Aug 2018, 00:52
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1
Thanos7 wrote:
What is the number of sides of a regular polygon in which 1/3rd of the sum of exterior angle is equal to the each interior angle ?

(A) 3

(B) 4

(C) 5

(D) 6

(E) 8



sum of interior angles = \((n-2)*180\)...... so each interior angle = \(\frac{(n-2)*180}{n}\).
Sum of exterior angles = 360

thus \(\frac{(n-2)*180}{n}=\frac{1}{3}*360.............(n-2)*180=120n..........180n-120n=360......n=6\).

D
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1) Absolute modulus : http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html
3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html


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Re: What is the number of sides of a regular polygon in which 1/3rd of the  [#permalink]

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New post 27 Aug 2018, 07:18
chetan2u wrote:
Thanos7 wrote:
What is the number of sides of a regular polygon in which 1/3rd of the sum of exterior angle is equal to the each interior angle ?

(A) 3

(B) 4

(C) 5

(D) 6

(E) 8



sum of interior angles = \((n-2)*180\)...... so each interior angle = \(\frac{(n-2)*180}{n}\).
Sum of exterior angles = 360

thus \(\frac{(n-2)*180}{n}=\frac{1}{3}*360.............(n-2)*180=120n..........180n-120n=360......n=6\).

D


chetan2u I may sound silly while asking this-
How to deduce the sum of exteriors=360?
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Re: What is the number of sides of a regular polygon in which 1/3rd of the  [#permalink]

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New post 27 Aug 2018, 07:37
1
siddreal wrote:
chetan2u wrote:
Thanos7 wrote:
What is the number of sides of a regular polygon in which 1/3rd of the sum of exterior angle is equal to the each interior angle ?

(A) 3

(B) 4

(C) 5

(D) 6

(E) 8



sum of interior angles = \((n-2)*180\)...... so each interior angle = \(\frac{(n-2)*180}{n}\).
Sum of exterior angles = 360

thus \(\frac{(n-2)*180}{n}=\frac{1}{3}*360.............(n-2)*180=120n..........180n-120n=360......n=6\).

D


chetan2u I may sound silly while asking this-
How to deduce the sum of exteriors=360?


Hi siddreal

Any polygon having n sides will have n angles and when you see each angle , the interior and exterior angles make a straight line and thus are equal to 180.

Therefore sum of all n interior angles and exterior angles is 180*n
But what is the sum of all interior angles of n sides = 180*(n-2)
So Sum of all exterior angles+180(n-2)=180n...
Sum of all exterior angles = 180n-180(n-2)=180n-180n+360=360
_________________

1) Absolute modulus : http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html
3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html


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Re: What is the number of sides of a regular polygon in which 1/3rd of the &nbs [#permalink] 27 Aug 2018, 07:37
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