If the sum of the interior angles of a polygon is at least 120 greater than that of its exterior angles, at last how many sides does the polygon have?
F.3 Maths question
2009-11-08 4:04 am
回答 (2)
2009-11-08 4:14 am
✔ 最佳答案
The sum of the exterior angles = 360The sum of the interior angles > 360 + 120 = 480
180 * (n - 2) > 480
n - 2 > 480/180 = 2.67
n > = 4.67
The polygon has at least 5 sides
2009-11-08 7:17 pm
I'm focus on the case that the interor angles of the polygon is equal to the sum of the exterior angles of the polygon +120.
Let the sides of the smallest polygon be n.
Since the sum of exterior angles of a polygon is 360.(By definition of the sum of exterior angles of polygon)
Therefore, the sum of the interior angles of the polygon =360+120=480
(n-2)*180>=480 (angles sum of poygon)
180n-360>=480
180n>=840
n>=14/3
Therefore , n at least must be 6.
Let the sides of the smallest polygon be n.
Since the sum of exterior angles of a polygon is 360.(By definition of the sum of exterior angles of polygon)
Therefore, the sum of the interior angles of the polygon =360+120=480
(n-2)*180>=480 (angles sum of poygon)
180n-360>=480
180n>=840
n>=14/3
Therefore , n at least must be 6.
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