For a convex polygon of n sides, we have:

i. Sum of all exterior angles = ...... .

ii. Sum of all interior angles = ...... .

iii. Number of diagonals = ...... .

i.

ii.

iii.

Exp:

(i) We will learn how to find the sum of the exterior angles of a polygon having n sides.

We know that,

Exterior angle + interior adjacent angle = 180°

So, if the polygon has n sides, then

Sum of all exterior angles + Sum of all interior angles = n × 180°

So, the sum of all exterior angles = n × 180° - Sum of all interior angles

Sum of all exterior angles = n × 180° - (n -2) × 180°

= n × 180° - n × 180° + 2 × 180°

= 180°n - 180°n + 360°

= 360°

Therefore, sum of all exterior angles of the polygon = 360°

(ii)

From any one of the vertices, say A₁, construct diagonals to other vertices.

There are altogether (n-2) triangles.

Sum of angles of each triangle = 180°

Sum of interior angles of an n-sided polygon

= (n-2) x 180°

(iii) Let there be a polygon ( I am assuming a convex polygon) of n sides.

Let us consider its one vertex. This does not form a diagonal with itself as well as immediate two neighbouring vertices,

hence it can form diagonals with remaining (n−3) vertices, (n−3) diagonals

Hence n vertices can form n(n−3) diagonals but ,

as each diagonal has been counted twice, as it was counted as fresh diagonal when either of the vertices at its ends was considered, but actually we have only one diagonal.

Hence the number of diagonals of a convex polygon of n sides is n(n−3)/2