Question

We know the sum of the interior angles of a triangle is 180°. Show that the sums of the interior angles of polygons with 3, 4, 5, 6, ... sides form an arithmetic progression. Find the sum of the interior angles for a 21 sided polygon.

Answer 1

Since, the sum of all interior angles of a polygon of n sides

= (2n – 4) × 90°

∴ Sum of interior angles of a polygon of sides 3

= (2 × 3 – 4) × 90°

= 180°

Sum of interior angles of a polygon of sides 4

= (2 × 4 – 4) × 90°

= 360°

Similarly, the sum of interior angles of the polygon of sides,

5, 6, 7 … are 540°, 720°, 900°…

Therefore, the series will be 180°, 360°, 540°, 720°, 900°… which is A.P.

Here a = 180°

d = 180°

We have to find the sum of interior angles of a polygon of 21 sides

i.e. 19^th term

a_n = a + (n – 1)d

a₁₉ = 180° + (19 – 1)180°

= 180° + 18 × 180°

= 180° + 3240°

= 3420°

Hence, the required sum of interior angles = 3420°.