The interioranglesof a polygon are in a.P. The smallest angle is 52and the common difference is 8. Find the number of sides of the polygon.Wages (in rs.)50-6060-7070-8080-9090-100100-110no. Of workers534p213
Answers
Let the number of sides in an polygon be n.
Therefore, sum of the Interior angle = (n - 2) × 180
Also, all the angles are in A.P., therefore,
a(First angle) = 52
Second angle = 52 + d, third angle = 52 + 2d, and so on. (where d is the difference = 8)
Therefore, sum of all angles in A.P. = n/2(2a + (n-1)d)
∴ Sn = n/2[2 × 52 + (n - 1)8]
Now, Both the sum is equal, therefore,
n/2[2 × 52 + (n - 1)8] = (n - 2)× 180
n[104 + 8n - 8] = (n - 2)360
n[96 + 8n] = 360[n - 2]
96n + 8n² = 360n - 720
8n² - 264n + 720 = 0
n² - 33n + 90 = 0
n² - 30n - 3n - 90 = 0
n(n - 30) - 3(n - 30) = 0
(n - 30)(n - 3) = 0
∴ n = 30 and n = 3
Hence, the no. of sides in an Polygon is either 3 or either 30.
Hope it helps.
Answer:
Let the number of sides in an polygon be n.
Therefore, sum of the Interior angle = (n - 2) × 180
Also, all the angles are in A.P., therefore,
a(First angle) = 52
Second angle = 52 + d, third angle = 52 + 2d, and so on. (where d is the difference = 8)
Therefore, sum of all angles in A.P. = n/2(2a + (n-1)d)
∴ Sn = n/2[2 × 52 + (n - 1)8]
Now, Both the sum is equal, therefore,
n/2[2 × 52 + (n - 1)8] = (n - 2)× 180
n[104 + 8n - 8] = (n - 2)360
n[96 + 8n] = 360[n - 2]
96n + 8n² = 360n - 720
8n² - 264n + 720 = 0
n² - 33n + 90 = 0
n² - 30n - 3n - 90 = 0
n(n - 30) - 3(n - 30) = 0
(n - 30)(n - 3) = 0
∴ n = 30 and n = 3