Question

exterior angle of a regular polygon having n-side is more than that of the polygon having n squre side by 50° find the number of the sideof each polygon

Answer 1

Exterior angle of a regular polygon having n side = 360/n

Exterior angle of a regular polygon having n² side = 360/n²

$\frac{360}{n} - \frac{360}{n^2} = 50 \\ \\ = \frac{360n^2 - 360n}{n^3} = 50 \\ \\ = \frac{360n(n - 1)}{n^3} = 50 \\ \\ = \frac{360n - 360}{n^2} = 50 \\ \\ \\ 360n - 360 = 50n^2 \\ \\ 50n^2 - 360n + 360 = 0 \\ \\ = 10(5n^2 - 36n + 36) = 0 \\ \\ \\ 5n^2 - 36n + 36 = \frac{0}{10} = 0$

$\\ \\ = 5n^2 - 30n - 6n + 36 = 0 \\ \\ = 5n(n - 6) - 6(n - 6) = 0 \\ \\ = (5n - 6)(n - 6) = 0 \\ \\ \\ 5n - 6 = 0; n = \frac{6}{5} \\ \\ n - 6 = 0; n = 6 \\ \\ \therefore n = 6$

6/5 is not a counting number, so it can't be considered as the no. of sides of the polygon. The value of n can be found through the quadratic equation formula too.

6 is the answer.

Let's check.

Exterior angle of a regular hexagon (6 sided) = 360/6 = 60°

Exterior angle of a regular triacontahexagon (6² = 36 sided or 36-gon) = 360/36 = 10°

60° - 10° = 50°

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Answer 2

It is hexagon and triacontahexagon with six and thirty six sides respectively.
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