A polygon with n sides inscribed in a circle has an angle sum of 180xn-360 So the problem is find n so that 180n-360=9000 => n=48 The regular polygon with an angle sum of 9000 has 48 sides
Im stuck on that too :/
The method of finding the sum of the interior angles of a polygon is by multiplying the (number of sides)-2 by 180, so the sum of the interior angle measures in a 25-sided polygon would be 23*180, or 4140 degrees.
Sum of interior angles of a 38 sided polygon: (38-2)*180 = 6480 degrees
The sum of the EXTERIOR angles of a polygon is 360 degrees. The sum of the INTERIOR angles of a polygon with n sides is (n-2)*180 degrees. If it is a regular polygon, each interior angle is (n-2)/n*180 degrees.
A polygon with n sides inscribed in a circle has an angle sum of 180xn-360 So the problem is find n so that 180n-360=9000 => n=48 The regular polygon with an angle sum of 9000 has 48 sides
Interior angles of any n-sided polygon sum to (2n - 4) right angles or 180n - 360 degrees. In your question this value = 9000, ie 180n = 9000 + 360 or 9360 degrees. 9360/180 = 52. A 52-sided polygon's interior angles total 9000 degrees.
Yes it is a 52-gon. (That means there are 52 sides)
There is no regular polygon with an interior angle of 20o. If the interior angle is 120o then there are six sides and the sum of the angles is 720o.
The sum of the interior angles of a polygon with n sides is 180*(n-2) degrees.
A 29-sided polygon.
A pentagon
No
The sum of each pair of interior and exterior angles of a polygon is always 180 degrees.
Im stuck on that too :/
360 ________________________________________________ Sum of Interior Angles of a Polygon = 180 (n-2) Sum of Exterior Angles of a Polygon = 360 The sum of an angle and an exterior angle of a regular polygon is 360
(number of sides -2)*180 = sum of interior angles