Question

The interior angle of a polygon Are in ap.if the smallest angle be 120 ans the common difference be 5 then find the number of sides

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 = 120° , common difference = 5°
∴ Sum of all interior angles of polygon
= n/2[2 * 120 + ( n – 1) * 5
Thus we should have
n/2 [2 * 120 + (n – 1) * 5] = (n – 2) * 180
⇒ n/2 [5n + 235] = (n – 2 ) * 180
⇒ 5n2 + 235n = 360n – 720
⇒ 5n2 – 125n + 720 = 0 ⇒ n2 – 25n + 144 = 0
⇒ (n – 16 ) (n – 9) = 0 ⇒ n = 16, 9
Also if n = 16 then 16th angle = 120 + 15 * 5 = 195° > 180°
∴ not possible.
Hence n = 9.
Thanks
Aditi Chauhan
askIITians Faculty
4 years ago
Smallest angle=120degreeCommon difference=5A P is 120, 125, 130,……..The sum of interior angles of a polygon= (n-2)180Hence Sum of n terms of an A P = (n-2)180n/2 {2.120+(n-1)5} = 180(n-2)5n^2 -125n +720 = 0n^2 -25n +144=0n=9 or 16hence number of sides can be 9 or 16