Q5 of 26 Page 54

The sum of the outer angles of polygon is twice the sum of the inner angles. How many sides does it have? What if the sum of outer angles is half the sum of inner angles? And if the sums are equal?

As we know the sum of exterior angles in each polygon is 360°.

Let n be the number of angles in the polygon.


Then the sum of interior angle is (n – 2) × 180°


Since the sum of interior angle is twice the sum of exterior angles


⇒ (n – 2) × 180° = 2 × 360°


⇒


⇒ (n – 2) = 2 × 2


⇒ (n – 2) = 4


⇒ n = 4 + 2


⇒ n = 6


Since the sum of exterior angle is half the sum of interior angles


⇒


⇒ (n – 2) × 180° = 2 × 360°


⇒


⇒ (n – 2) = 2 × 2


⇒ (n – 2) = 4


⇒ n = 4 + 2


⇒ n = 6


Since the sum of interior angle is equal to the sum of exterior angles


⇒ (n – 2) × 180° = 360°


⇒


⇒ n – 2 = 2


⇒ n = 2 + 2


⇒ n = 4


More from this chapter

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3

Draw a quadrilateral and mark any two outer angles. Is there any relation between the sum of these two and the inner angles at the outer two and the inner angles at the other two vertices?

 4

In a polygon with all angles equal, one outer angle is twice an inner angle.

i) How much is each of its angle?


ii) How many sides does it have?

 1

Draw a hexagon of equal sides and unequal angles.

 2

Draw a hexagon of equal angles and unequal sides.