Question

Exterior angle of a regular polygon having n sides is more than
that of the polygon having n^(n square) sides by 50°. Find the number of
the sides of each polygon.​

Answer 1

Correct Question :-

Exterior angle of a regular polygon having n sides is more than that of the polygon having n² sides by 50°. Find the number of the sides of each polygon.

Solution :-

Let the number of sides of the regular polygon be 'n'

By Exterior angle formula

Exterior angle of regular polygon having n sides = 360°/n

Exterior angle of the polygon having n² sides = 360°/n²

Given :-

Exterior angle of a regular polygon = 50 more than the Exterior angle of polygon havin n² sides

$\implies \dfrac{360^{ \circ} }{n} = \dfrac{360^{ \circ} }{ {n}^{2} } + 50^{ \circ}$

$\implies 0 = \dfrac{360 }{ {n}^{2} } - \dfrac{360}{n} + 50$

$\implies 0 = \dfrac{360 - 360n + 50 {n}^{2} }{ {n}^{2} }$

$\implies 50 {n}^{2} - 360n + 360 = 0$

$\implies 5 {n}^{2} - 36n + 36 = 0$

$\implies 5 {n}^{2} - 30n - 6n+ 36 = 0$

$\implies 5n(n - 6) - 6(n - 36) = 0$

$\implies (5n - 6)(n - 6) = 0$

$\implies 5n - 6 = 0 \qquad n - 6= 0$

$\implies 5n = 6 \qquad n= 6$

$\implies n = \dfrac{6}{5} \qquad n= 6$

Number of sides can't be a fraction

$\implies n= 6$

Number of sides of regular polygon = n = 6

Number sides of another polygon = n² = 6² = 36

Therefore 6 and 36 are the number of sides of each polygon.