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In pentagon ABCDE above, x =

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In pentagon ABCDE above, x =  [#permalink]

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New post 11 Jan 2019, 01:17
00:00
A
B
C
D
E

Difficulty:

  25% (medium)

Question Stats:

89% (01:03) correct 11% (02:01) wrong based on 18 sessions

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Re: In pentagon ABCDE above, x =  [#permalink]

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New post 11 Jan 2019, 04:02
Sum of the interior angles of a pentagon = 18(n-2); where n=5 > 180(3) = 540

x = (540 - (180+130))/2 = 115

Answer: E

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Re: In pentagon ABCDE above, x =  [#permalink]

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New post 11 Jan 2019, 04:12
Bunuel wrote:
Image
In pentagon ABCDE above, x =

A. 65
B. 75
C. 80
D. 105
E. 115


Attachment:
2019-01-11_1315.png


sum of internal angle of pentagon = (n-2) * 180 = 3 * 180 = 540
sum ; 2x+130+180 = 540
x= 115
IMO E
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Re: In pentagon ABCDE above, x =  [#permalink]

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New post 11 Jan 2019, 10:41
Sum of angles in a polygon of n sides= (n-2)*180
Here, pentagon has n=5
So, 90+90+130+x+x=(5-2)*180=540
Hence, x=115
GMAT Club Bot
Re: In pentagon ABCDE above, x = &nbs [#permalink] 11 Jan 2019, 10:41
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In pentagon ABCDE above, x =

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