We know that the sum of interior angles of a triangle is 180°. Show that the sums of interior angles of polygons with 3, 4, 5, 6 … sides from an arithmetic progression. Find the sum of interior angles for a 21 sided polygon. [NCERT EXEMPLAR]
Given: the sum of interior angles of a triangle is 180°
To prove: Sum of interior angles of polygons with 3, 4, 5,…… forms an A.P.
To find: Sum of interior angles for a 21 sided polygon i.e. a21
We know,
The sum of interior angles of a polygon with n sides,
an = 180° × (n – 2)
Sum of interior angles of a polygon of 3 sides,
a3= 180° × (3 – 2) = 180° × 1 = 180°
Sum of interior angles of a polygon of 4 sides,
a4= 180° × (4 – 2) = 180° × 2 = 360°
Sum of interior angles of a polygon of 5 sides,
a5= 180° × (5 – 2) = 180° × 3 = 540°
A.P is known for Arithmetic Progression whose common difference, d = an – an-1 where n > 0
Here,
d = a4 – a3 = 360° - 180° = 180°
d = a5 – a4 = 540° - 360° = 180°
Common difference is same in both the cases
This shows that sum of interior angles of polygons with 3, 4, 5,…… forms an A.P.
Hence Proved
Sum of interior angles of a polygon of 21 sides,
a21= 180° × (21 – 2) = 180° × 19 = 3420°