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In a particular seven-sided polygon, the sum of four equal
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21 Mar 2019, 10:49
21 Mar 2019, 10:49
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Question Stats:
90% (01:28) correct
9% (02:14) wrong based on 43 sessions
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In a particular seven-sided polygon, the sum of four equal interior angles, each equal to \(a\) degrees, is equivalent to the sum of the remaining three interior angles.
Quantity A is greater.
Quantity B is greater.
The two quantities are equal.
The relationship cannot be determined from the information given.
Quantity A |
Quantity B |
\(a°\) |
\(110°\) |
Quantity A is greater.
Quantity B is greater.
The two quantities are equal.
The relationship cannot be determined from the information given.
Retired Moderator
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Re: In a particular seven-sided polygon, the sum of four equal
[#permalink]
23 Mar 2019, 08:24
23 Mar 2019, 08:24
1
Carcass wrote:
In a particular seven-sided polygon, the sum of four equal interior angles, each equal to \(a\) degrees, is equivalent to the sum of the remaining three interior angles.
Quantity A is greater.
Quantity B is greater.
The two quantities are equal.
The relationship cannot be determined from the information given.
Quantity A |
Quantity B |
\(a°\) |
\(110°\) |
Quantity A is greater.
Quantity B is greater.
The two quantities are equal.
The relationship cannot be determined from the information given.
Useful rule: the sum of the angles in an n-sided polygon = (n - 2)(180°)
So, the sum of the angles in a 7-sided polygon = (7 - 2)(180°) = (5)(180°) = 900°
The sum of four equal interior angles, each equal to a degrees, is equivalent to the sum of the remaining three interior angles.
The sum of the 4 equal angles = 4a°
So, the sum of the 3 REMAINING angles = 900° - 4a° [since the sum of all 7 angles is 900°]
So, we can write: 4a° = 900° - 4a°
Add 4a° to both sides to get: 8a° = 900°
Solve: a = 900/8 = 112.5°
We get:
QUANTITY A: 112.5°
QUANTITY B: 110°
Answer: A
Cheers,
Brent
gmatclubot
Re: In a particular seven-sided polygon, the sum of four equal [#permalink]
23 Mar 2019, 08:24
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