SUM OF THE MEASURES OF THE EXTERIOR ANGLES OF A POLYGON

Understanding quadrilaterals of Class 8

THEOREM TITLE:
(Exterior angle sum property) If the sides of a quadrilateral are produced in order, the sum of four exterior angles so formed is 360º.

Proof:
Let the sides of a quadrilateral be produced in order as shown in figure, forming exterior angles ∠1, ∠2, ∠3 and ∠4.
Since ∠1 and ∠ form a linear pair and the sum of the angles of a linear pair is 180º.
∴∠1 + ∠A = 180º ----(i)
Similarly, we have
∠2 + ∠B = 180º ----(ii)
∠3 + ∠C = 180º ----(iii)
and, ∠4 + ∠D = 180º ----(iv)
quadrilateral
Adding equation (i) to (iv), we have
(∠1 + ∠2 + ∠3 + ∠4) + (∠A + ∠B + ∠C + ∠D) =  180º + 180º + 180º + 180º
⇒ ∠1 + ∠2 + ∠3 + ∠4 + 360º = 720º  [∠A + ∠B + ∠C + ∠D = 360º]
→ ∠1 + ∠2 + ∠3 + ∠4 = 720º - 360º = 360º


THEOREM TITLE:


The sum of all the exterior angles formed by producing the sides of a convex polygon in the same order is equal to four right angles.
Given: A convex polygon P1 P2 P3 P4 P5.Its sides P1 P2,P2P3,P3P4, P5P1are produced in order, forming exterior angles ∠1, ∠2, ∠3, ∠4 and ∠5.
To Prove: ∠1 + ∠2 + ∠3 + ∠4 + ∠5 = 4 right angles.
Construction: Take any point O, outside the polygon. Draw OA1,OA2,OA3,OA4, and OA5 parallel to and in the same sense as P1P2,P3P4,P4 P5 and P4 P5, and P5 P1 respectively.
convex polygon

 

Proof: Since the arms of ∠ and ∠a are parallel and drawn in the same sense. ∴ ∠1 = ∠a
Similarly,∠2 = ∠b, ∠3 = ∠c,∠4 = ∠d and ∠5 = ∠e
∴ ∠1 + ∠2 + ∠3 + ∠4 + ∠5 = ∠a + ∠b + ∠c + ∠d + ∠e = 360º [Sum of the angles at a point is 360º]
= 4 x 90º
= 4 right angles.


note ➲Each exterior angle of a regular polygon of n sides is equal to 360 degree/n
➲If there is a regular polygon of n sides (n ≥ 3), then its each interior angle is equal to (2n-4)/nright angles i.e.,((2n-4)/n)90

 

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