Question

To verify by paper cutting and pasting that the sum of the exterior angles drawn in order, of any polygon is 360°

Answer 1

Answer:

Let sum of all exterior angles be 'E', and sum of all interior angles be 'I'. E = n × 180° - (n -2) × 180°. Hence, The sum of all the exterior angles of a polygon is 360° .

Step-by-step explanation:

Answer 2

Answer:

A polygon with ‘n’ number of sides, lets say <1, <2, <3, <4, <5,…..<n are the exterior angles and lets at that A, B, C, D, E….n are interior angles.

To prove:

<1 + <2 + <3 + <4 + <5 + …. + <n = 360 degrees.

Solution :

We know that the sum of interior angles of any given polygon is = (n-2) x 180 degrees

Therefore, we can say that, <1 + <A = 180

<2 + <B = 180

<3 + <C = 180

<4 + <D = 180

<5 + <E = 180

And this goes on upto n times.

By adding the above equations, we get :

<1 +<2+<3+<4+<5+…+<n = 180n - (A+B+C+D+E+…+n)

= 180n - { (n-2) x 180 }

= 180n - 180n + ( 2x 180 )

= 2 x 180

= 360 degrees

Therefore,

<1 +<2+<3+<4+<5+…+<n = 360 degrees.

Hence proved.

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