Algebra> The interior angles of a polygon are in a...

The interior angles of a polygon are in arithmetic progression. The smallest angle is 120°, and the common difference is 5°, Find the number of sides of the polygon.

Simran Bhatia , 10 Years ago

Grade 11

3 Answers

Aditi Chauhan

Last Activity: 10 Years ago

Hello Student,

Please find the answer to your question

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 16^th angle = 120 + 15 * 5 = 195° > 180°

∴ not possible.

Hence n = 9.

Thanks
Aditi Chauhan
askIITians Faculty

Vivek Kumar

Last Activity: 7 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