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the sum of the interior angles of the polygon above
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30 Mar 2021, 04:10
30 Mar 2021, 04:10
Expert Reply
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Question Stats:
85% (00:39) correct
14% (01:02) wrong based on 21 sessions
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Attachment:
GRE pentagon.jpg [ 13.84 KiB | Viewed 1688 times ]
Quantity A |
Quantity B |
6b |
the sum of the interior angles of the polygon above |
A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.
Kudos for the right answer and explanation
Question part of the project GRE Quantitative Reasoning Daily Challenge - (2021) EDITION
GRE - Math Book
Re: the sum of the interior angles of the polygon above
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30 Mar 2021, 07:45
30 Mar 2021, 07:45
1
Qt A
From figure 1, we can deduce b = 80
6b = 6 * 80 = 480
Qt B
Sum of interior angle of polygon = 180 ( n - 2 ) (n = no of angle )
sum = 180 ( 5-2 ) = 540
Qt B > Qt A
Answer B
From figure 1, we can deduce b = 80
6b = 6 * 80 = 480
Qt B
Sum of interior angle of polygon = 180 ( n - 2 ) (n = no of angle )
sum = 180 ( 5-2 ) = 540
Qt B > Qt A
Answer B
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Re: the sum of the interior angles of the polygon above
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30 Mar 2021, 07:59
30 Mar 2021, 07:59
1
Carcass wrote:
Attachment:
GRE pentagon.jpg
Quantity A |
Quantity B |
6b |
the sum of the interior angles of the polygon above |
A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.
\(b = 80\) ..... (Vertically opposite angles are equal)
Sum of all interior angles in a Polygon = \((n - 2)(180)\), where \(n\) is number of sides
Col. A: \(6(80) = 480\)
Col. B: \((5 - 2)(180) = 540\)
Hence, option B
