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The sum of the interior angles of any polygon with n sides is 180(n -

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HKD1710
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The sum of the interior angles of any polygon with n sides is 180(n - [#permalink] The sum of the interior angles of any polygon with n sides is 180(n -   23 Dec 2018, 09:44
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The sum of the interior angles of any polygon with n sides is 180(n - 2) degrees. If the sum of the interior angles of polygon P is three times the sum of the interior angles of quadrilateral Q, how many sides does P have?

(A) 6
(B) 8
(C) 10
(D) 12
(E) 14
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B
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sahibr
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Re: The sum of the interior angles of any polygon with n sides is 180(n - [#permalink] The sum of the interior angles of any polygon with n sides is 180(n -   23 Dec 2018, 13:57
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Answer is B ie. 8.

Sum of the internal angles of a quadrilateral is 360.
{Even if during the exam, the above fact isn't known by the examinee, one can simply substitute 4 in the formula stated in the question to get 360. 180(n-2) => 180(4-2) => 360}


Question states that sum of internal angles of a certain polygon P is three times the sum of internal angles of quadrilateral Q.

Hence, the Sum of internal angles(S) = 3 x 360 = 1080.

S=1080=180(n-2)

Hence, n=8
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The sum of the interior angles of any polygon with n sides is 180(n - [#permalink] The sum of the interior angles of any polygon with n sides is 180(n -   23 Dec 2018, 18:53
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180(n-2)=3*360

n= 8

IMO B



HKD1710 wrote:
The sum of the interior angles of any polygon with n sides is 180(n - 2) degrees. If the sum of the interior angles of polygon P is three times the sum of the interior angles of quadrilateral Q, how many sides does P have?

(A) 6
(B) 8
(C) 10
(D) 12
(E) 14
 

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Re: The sum of the interior angles of any polygon with n sides is 180(n - [#permalink] The sum of the interior angles of any polygon with n sides is 180(n -   24 Dec 2018, 01:12
By definition, a quadrilateral has 4 sides and the sum of interior angles is 360.

From the question stem, we have been told that the sum of the interior angles of
the polygon is thrice the sum of the interior angles of the polygon(\(3*360 = 1080\))

Solving for n, we get \(180(n - 2) = 1080\) -> \(n - 2 = \frac{1080}{180} = 6\) -> \(n = 8\)

Therefore, Polygon P has 8(Option B) sides whose interior angles sum thrice that of a quadrilateral.
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Re: The sum of the interior angles of any polygon with n sides is 180(n - [#permalink] The sum of the interior angles of any polygon with n sides is 180(n -   10 May 2024, 17:25
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