Question

exterior angle of a regular polygon having n sides is more than that of the polygon having n2 sides by 50" find no. of sides​

Answer 1

Question :-- exterior angle of a regular polygon having n sides is more than that of the polygon having n² sides by 50 .. find no. of sides ?

Formula used :---

→ Each exterior angle of Regular polygon is given by 360°/n..

Solution :---

According to Question we can say that,

→ (360°/n) - (360/n²) = 50°

Taking LCM of Denominator in LHS, we get,

→ (360n - 360) / n² = 50

Cross - Multiply

→ (360n - 360) = 50n²

→ 10(36n - 36) = 50n²

Dividing both sides by 10 now, we get,

→ 36n - 36 = 5n²

→ 5n² - 36n + 36 = 0

Splitting the middle term now,

→ 5n² - 30n - 6n + 36 = 0

→ 5n(n-6) -6(n-6) = 0

→ (n-6)(5n-6) = 0

Putting both Equal to zero now,

→ n - 6 = 0 or, 5n-6 = 0

→ n = 6 . or , 5n = 6, => n = 6/5 .

Hence, Number of sides of regular polygon will be 6.

Answer 2

$\huge{\bf{Solution}}$

According to Question we can say that,

→ (360°/n) - (360/n²) = 50°

Taking LCM of Denominator in LHS, we get,

→ (360n - 360) / n² = 50

Cross - Multiply

→ (360n - 360) = 50n²

→ 10(36n - 36) = 50n²

Dividing both sides by 10 now, we get,

→ 36n - 36 = 5n²

→ 5n² - 36n + 36 = 0

Splitting the middle term now,

→ 5n² - 30n - 6n + 36 = 0

→ 5n(n-6) -6(n-6) = 0

→ (n-6)(5n-6) = 0

Putting both Equal to zero now,

→ n - 6 = 0 or, 5n-6 = 0

→ n = 6 . or , 5n = 6, => n = 6/5 .

Hence, Number of sides of regular polygon will be 6.