Qstnid -B-15- The Difference Between The Exterior Angles Of Two Regular Polygons, Having The Sides Equal To (N - 1) And (N + 1) Is 9°. Find The Value Of N. - 16: Understanding Shapes (Including Polygons) Class 8 Maths - Icse-viii-mathematics

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ICSE-VIII-Mathematics

16: Understanding Shapes (Including Polygons) Class 8 Maths

with Solutions - page 5

Qstn# B-15 Prvs-Qstn Next-Qstn

#15
The difference between the exterior angles of two regular polygons, having the sides equal to (n - 1) and (n + 1) is 9°. Find the value of n.
Ans : We know that sum of exterior angles of a polynomial is 360°
If sides of a regular polygon = n - 1
Then each angle = 360˚/(n - 1)
And if sides are n + 1, then each angle = 360˚/(n +1)
According to the condition,
360˚/(n - 1) - 360˚/(n + 1) = 9
⇒ 360˚[1/(x - 1) - 1/(x + 1)] = 9
⇒ 360˚ [(n + 1 - n + 1)/(n - 1)(n + 1)] = 9
⇒ (2 × 360)/(n² - 1) = 9
⇒ n² - 1 = (2×360)/9 = 80
⇒ n² - 1 = 80
⇒ n² = 1 - 80 = 0
⇒ n² - 81 = 0
⇒ (n)² - (9)² = 0
⇒ (n + 9)(n - 9) = 0
Either n + 9 = 0, then n = -9 which is not possible being negative,
Or, n - 9 = 0, then n = 9
∴ n = 9
∴ No. of sides of a regular polygon = 9