Question

The interior angles of a polygon are in A.P. If the smallest angle is 100° and the common difference is 4°, then find the no of sides of the polygon

Answer 1

Let there be in n sides in the polygon.

Then by geometry, sum of all n interior angles of polygon = (n – 2) * 180°

Also the angles are in A. P. with the smallest angle = 100° , common difference = 4°

∴ Sum of all interior angles of polygon

= n/2[2 * 100 + ( n – 1) * 4]

Thus we should have

n/2 [2 * 100 + (n – 1) * 4] = (n – 2) * 180

⇒ n/2 [4n + 196] = (n – 2 ) * 180

⇒ 4n*n+ 196n = 360n – 720

⇒ 4n*n – 164n + 720 = 0 ⇒ n*n– 41n + 180= 0

⇒ (n – 36) (n – 5) = 0 ⇒ n = 36, 5

Also if n = 36 then 36th angle = 100 + 35 * 4 = 240° > 180°

∴ not possible.

Hence n = 5.