To find the sum of the interior angles of a 50-gon (a polygon with 50 sides), you can use the formula: ( \text{Sum of interior angles} = (n - 2) \times 180^\circ ), where ( n ) is the number of sides. For a 50-gon, this would be ( (50 - 2) \times 180^\circ = 48 \times 180^\circ = 8640^\circ ). Thus, the sum of the interior angles of a 50-gon is 8640 degrees.
There are 48 triangles in a 50-gon polygon
Sum of what aspect of a 20-gon: the length of its sides, its interior angles, its exterior angles, its apothems?
The interior angles of a 30-gon add up to 5040 degrees
hahaa.... i don't know!
1620 degrees
The sum of the interior angles of a n-gon is (n-2)*180 degrees (n>2). So the sum for a 50-gon would be 48*180 = 8640 degrees.
9000
The sum of the exterior angles of a polgon - no matter how many sides - is 360 degrees.
You multiply it by 180 degrees. So 180x50=9000
The sum of interior angle ( Sum-of-interior-angles-n-gon) of n-gon is:If we have a ,Then
8640 degrees The sum of the interior angles of an n-sided polygon is: 180(n-2). For n=50, we have 180(50-2)=180(48)=8640 degrees
the sum of the interior angles of a 35-gon.
There are 48 triangles in a 50-gon polygon
Enneacontagon
the sum of the interior angles of a convex 45-gon is 7740.
Sum of what aspect of a 20-gon: the length of its sides, its interior angles, its exterior angles, its apothems?
180x28 is the sum of the interior angles in a 30- gon