Tag: maths

Given, regular polygon whose each exterior angle is 40% of a right angle = $ 90^o \times \dfrac{40}{100} = 36^o $
Sum of all exterior angle of any polygon is $ 360^o $
Now,
$ 36^o \times$  number    of   angles  = $ 360^o $
The number  of  angles =$ 10 $
Any polygon have equal number of angles and sides.
Therefore the number of side of the polygon is 10.
Since the number of sides is an integer, therefore their exist a polygon whose each exterior angle is 40% of a right angle. 

Given, a regular polygon whose each exterior angle is $32^o $
Each exterior angle of a regular polygon = $ \dfrac {360^o}{n} $, where n = number of side
Now,
$ \dfrac {360^o}{n} = 32^o $
$=> n = \dfrac{45}{4} $
Since, n should be an integer, so it is not possible a regular polygon whose each exterior angle is $32^o $

Let the number of sides of the polygon is n. (which must be an integer)
If each exterior angle is $ 80^o $, then sum of all exterior angle is $ n \times 80^o $.
And Sum of all exterior angles = $ 180^o $
$=>  n \times 80^o = 180^o $
$=> n = 1.25 $
So, it is not possible to have a regular polygon whose each exterior angle is $ 80^o $

Given, a regular polygon whose each exterior angle is $20^o $
Each exterior angle of a regular polygon = $ \dfrac {360^o}{n} $, where n = number of side
Now,
$ \dfrac {360^o}{n} = 20^o $
$=> n = 18 $
Since, n should be an integer, so their exist a regular polygon whose each exterior angle is $20^o $

The sum of the exterior angles of a regular polygon is $360^o$

Each exterior angle of a regular polygon of 9 sides
$=\dfrac {360^o}{n}$, where $n=9=\left (\dfrac {360^o}{9}\right )^o=40^o$

If n is the number of sides of the polygon, then $(2n -4)\times 90^o = 1440$
or 2n = 20          or          n = 10

Given the exterior angle of regular polygon is 12

$\Rightarrow$ Sum of exterior angles of a regular hexagon $=360^o$