Question

An exterior angle of a regular polygon having n-sides is more than that of the polygon having $n^2$ sides by ${50}^{\circ }$, then the number of the sides of each polygon are 6 and 38.

State true or false.

• A. True
• B. False

Answer 1

The correct option is B False

Given: An exterior angle of a regular polygon having $n$-sides is more than that of the polygon having $n^2$ sides by ${50}^{\circ }$

To find the number of the sides of each polygon are 6 and 38.

Solution:

We know in any polygon with $n$ number of sides, the exterior angle, $A=\frac{{360}^{\circ }}{n}$

Hence according to the given conditions,

$\frac{{360}^{\circ }}{n}=\frac{{360}^{\circ }}{n^2}+{50}^{\circ }⟹{360}^{\circ }n=({50}^{\circ }{n}^2+{360}^{\circ })⟹5n^2-36n+36=0$

This is of the form $a{x}^2+bx+c=0$ where $a=5,b=-36,c=36$, hence the roots of the equation will be

$n=\frac{-b\pm \sqrt{b^2-4ac}}{2a}⟹n=\frac{36\pm \sqrt{(36{)}^2-4(5\left)\right(36)}}{2\left(5\right)}⟹n=\frac{36\pm \sqrt{1296-720}}{10}⟹n=\frac{36\pm \sqrt{576}}{10}⟹n=\frac{36\pm 24}{10}$

or $n=\frac{36+24}{10},\frac{36-24}{10}⟹n=6,1.2$

As the sides cannot be in decimal. So $n=6$

Hence the number of the sides of each polygon are $n=6$ and $n^2=36$