Question

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

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{know}\mathrm{that}\mathrm{each}\mathrm{exterior}\mathrm{angle}\mathrm{of}\mathrm{a}\mathrm{regular}\mathrm{polygon}\text{of}n\mathrm{side}s\mathrm{is}\left(\frac{360}{n}\right)° \\ \mathrm{Thus},\mathrm{each}\mathrm{exterior}\mathrm{angle}\mathrm{of}\mathrm{a}\mathrm{regu}\mathrm{lar}\mathrm{polygon}of{n}^2\mathrm{sides}\mathrm{is}\left(\frac{360}{n^2}\right)° \\ \mathrm{By}\mathrm{the}\mathrm{given}\mathrm{condition},\mathrm{we}\mathrm{get}: \\ \left(\frac{360}{n}\right)°=\left(\frac{360}{n^2}\right)°+50° \\ \Rightarrow \frac{360}{n^2}-\frac{360}{n}+50=0 \\ \Rightarrow \frac{360-360n+50n^2}{n^2}=0 \\ \Rightarrow 50n^2-360n+360=0 \\ \Rightarrow 10\left(5n^2-36n+36\right)=0 \\ \Rightarrow 5n^2-36n+36=0 \\ \mathrm{On}\mathrm{splititng}\mathrm{the}\mathrm{middle}\mathrm{term}-36n\mathrm{as}-30n-6n,\mathrm{we}\mathrm{get}: \\ 5n^2-30n-6n+36=0 \\ \Rightarrow 5n\left(n-6\right)-6\left(n-6\right)=0 \\ \Rightarrow \left(n-6\right)\left(5n-6\right)=0 \\ \Rightarrow n-6=0or5n-6=0 \\ \Rightarrow n=6orn=\frac{6}{5} \\ \mathrm{Since}n\mathrm{is}\mathrm{the}\mathrm{number}\mathrm{of}\mathrm{side}so\mathrm{f}\mathrm{the}\mathrm{polygon},n=6 \\ \mathrm{Then},\mathrm{the}\mathrm{polygon}\mathrm{with}\mathrm{n}\mathrm{sides}\mathrm{has}6\mathrm{sides}\mathrm{and}\mathrm{that}\mathrm{with}{\mathrm{n}}^2\mathrm{sides}\mathrm{has}36\mathrm{sides}. \end{gathered}$$