Question

What is the sum of the exterior angles of a polygon with 4 sides? 5 sides? 6 sides? N sides?

Answer 1

Hint: A polygon is a figure formed by 3 or more than 3 line segments. An exterior angle is the angle formed by any one of the sides of any closed figure like a polygon by extending its adjacent side.
Formulas used: Interior angle $ = \dfrac{{\left( {n - 2} \right) \times 180^\circ }}{n} $ and Exterior angle $ = 180 - \dfrac{{\left( {n - 2} \right)180}}{n} $ .

Complete step by step solution:
In this question, we are supposed to find the sum of exterior angles of a polygon with 4 sides, 5 sides, 6 sides, and n sides.
First of all, polygon is a closed figure formed by 3 or more than 3 line segments.
Interior angle is the angle made by two adjacent sides inside the closed figure and exterior angle means the angle made by two adjacent sides by extending them outside the figure.
Now, the sum of interior angles of a polygon is given by the formula
The sum of interior angles $ = \left( {n - 2} \right) \times 180^\circ $
And each angle $ = \dfrac{{\left( {n - 2} \right) \times 180^\circ }}{n} $ - - - - - - - - (1)
Now, the sum of exterior angle and an interior angle is equal to $ 180^\circ $ .
Therefore, exterior angle $ = 180 - $ interior angle.
Now, from equation (1),
Exterior angle $ = 180 - \dfrac{{\left( {n - 2} \right)180}}{n} $ - - - - - - - - (2)
Now, we have all the required formulas.
Let’s solve our question.

Case 1: 4 sides
For $ n = 4 $ ,
From equation (2),
Exterior angle $ = 180 - \dfrac{{\left( {4 - 2} \right)180}}{4} $
 $
   = 180 - \dfrac{{360}}{4} \\
   = 180 - 90 \\
   = 90 \;
  $
Therefore, sum of all exterior angles $ = 4 \times 90 $
 $ = 360 $

Case 2: 5 sides
For $ n = 5 $ ,
From equation (2),
Exterior angle $ = 180 - \dfrac{{\left( {5 - 2} \right)180}}{5} $
 $
   = 180 - \dfrac{{3 \times 180}}{5} \\
   = 180 - 108 \\
   = 72 \;
  $
Therefore, sum of all exterior angles $ = 5 \times 72 $
 $ = 360 $

Case 3: 6 sides
For $ n = 6 $ ,
From equation (2),
Exterior angle $ = 180 - \dfrac{{\left( {6 - 2} \right)180}}{6} $
 $
   = 180 - \dfrac{{4 \times 180}}{6} \\
   = 180 - 120 \\
   = 60 \;
  $
Therefore, sum of all exterior angles $ = 6 \times 60 $
 $ = 360 $

Case 4: n sides
For n sides,
From equation (2),
Exterior angle $ = 180 - \dfrac{{\left( {n - 2} \right)180}}{n} $
Taking L.C.M, we get
Exterior angle\[ = \dfrac{{180n - 180n + 360}}{n}\]
 $ = \dfrac{{360}}{n} $
Therefore, sum of all exterior angles $ = n \times \dfrac{{360}}{n} $
 $ = 360 $

Note: Notice that in every case, the sum of exterior angles is 360. It is a property of polygons that the sum of exterior angles is always 360. Exterior angle $ = 180 - \dfrac{{\left( {n - 2} \right)180}}{n} $ where n is equal to the side of the polygon.