Question

The angle in one regular polygon is to that in another as 3 : 2 and the number of sides in first is twice that in the second. Determine the number of sides of two polygons.

Answer 1

Let the number of sides in the first polygon be 2x and the number of sides in the second polygon is x.
We know:
Angle of an n-sided regular polygon = $\left(\frac{n-2}{n}\right)\mathrm{\pi }$ radian
∴ Angle of the first polygon = $\left(\frac{2x-2}{2x}\right)\mathrm{\pi }=\left(\frac{\mathrm{x}-1}{\mathrm{x}}\right)\mathrm{\pi }$ radian

Angle of the second polygon = $\left(\frac{x-2}{x}\right)\mathrm{\pi }$ radian
Thus, we have:
$$\begin{gathered} \frac{\left({\displaystyle \frac{\mathrm{x}-1}{\mathrm{x}}}\right)\mathrm{\pi }}{\left({\displaystyle \frac{\mathrm{x}-2}{\mathrm{x}}}\right)\mathrm{\pi }}=\frac{3}{2} \\ \Rightarrow \frac{x-1}{x-2}=\frac{3}{2} \\ \Rightarrow 2x-2=3x-6 \\ \Rightarrow x=4 \end{gathered}$$
Thus,
Number of sides in the first polygon = 2x = 8
Number of sides in the first polygon = x = 4