3.4 Sum of exterior angles of a Polygon
Figure 3.4
Figure 3.4 shows a pentagon. Its external angles are named from a to e. The aim is to find the sum of these five angles.
It is known that the sum of internal angles of a Pentagon
= (5  2 )
´ 180^{0}
= 3 ´
180^{0}
= 54^{0}
\ each interior
angle of the pentagon measures
540^{0} /
5 = 108^{0}
The interior and exterior angles form linear pairs and hence are supplementary.
\ Each exterior
angle measures 180^{0}  108^{0} = 72^{0}
\ Sum
of five exterior angles = 5 ´
72 = 360^{0}
It can be proved that the sum of the exterior angles for any polygon is 360 ^{0.}
Sum of interior angles of an n sided polygon = ( n  2 ) 180^{0}.
\ Measure of
each internal angle =
\ Each exterior
angle =
\ Sum of n exterior
angles =
=
=
Conclusion : The sum of interior angles of a polygon
is dependent on the number of sides but the sum of the exterior
angles is always 360^{0}.
