Angles of polygon are in the arithmetic sequence 120,125,130,............calculate number of sides of the polygon
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Hey mate here is your answer ⤵⤵⤵⤵
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
⇒ 5n^2 + 235n = 360n – 720
⇒ 5n^2 – 125n + 720 = 0 ⇒ n^2 – 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.
❤hope it helps you ❤
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
⇒ 5n^2 + 235n = 360n – 720
⇒ 5n^2 – 125n + 720 = 0 ⇒ n^2 – 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.
❤hope it helps you ❤
ajithkumarbindhu67:
I cant understand this
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