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23 Dec 2018, 09:44
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
15%
(low)
Question Stats:
81% (01:30) correct
19%
(01:50)
wrong
based on 672
sessions
History
Date
Time
Result
Not Attempted Yet
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
(A) 6
(B) 8
(C) 10
(D) 12
(E) 14
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
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
General Discussion
Archit3110
23 Dec 2018, 18:53
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180(n-2)=3*360
n= 8
IMO B
n= 8
IMO B
HKD1710
