# If the exterior angle of a regular polygon is 45 degrees, then what is the no. of sides in the polygon and the no. of diagonals?I need this answer with the steps and formulas used

2
by Sam1142

2014-10-07T19:37:58+05:30
sum of exterior angles = 360 degrees

no of sides(n) = sum of exterior angles of a regular polygon / given exterior angle of the regular polygon
⇒360/45
⇒8 sides.
no of diagonals= n(n-3)/2⇒8(8-3)/2⇒8(5)/2⇒40/2⇒20 diagonals
Therefore, number of sides are 8 and number of diagonals are 20
lets see
Comment has been deleted
because der shuld be at least 2 ans.
Comment has been deleted
ok
2014-10-08T12:37:20+05:30

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A regular polygon has n sides and n vertices. The lines joining the vertices and the center O of polygon create n isosceles triangles. The side of the polygon becomes base of these triangles. The angle at the center in the triangle is Ф = 360°/n.

As the two angles at the base are A/2 = (180° - Ф )/2.
The interior angle at a vertex is A = 180
°Ф.

So the exterior angle  is =
Ф = 360°/n = 2π/n
So if 360°/n = 45°,           n  = 360°/45 = 8

It is a regular octagon with 8 sides.

The number of line connecting each vertex to another is :  C = 8 * 7 /2  = 28.

Of these, there are 8 sides among adjacent vertices.

The remaining are the diagonals and are 28 – 8 = 20.

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n >= 3

Formula for number of diagonals =  n(n-1)/2  - n = n(n-3)/2.

Sum total of all exterior angles = 360° = 2 π  for any regular  polygon.

One exterior angle = 360° / n = 2π/n
one interior angle = 180° – 360°/n = 180° (n-2)/n  =  (n-2)π/n
Angle made by a side at the cente = 360°/n  = 2π/n
Sum total of all interior angles = n * 180° (n-2)/n  = 180° (n-2)   = (n-2)π