The area of a regular polygon of n-sides inscribed in a circle is to that of the same number of sides circumscribing the same circle is 3:4 then n=

$$\begin{gathered} \mathrm{Area} \mathrm{of} \mathrm{a} \mathrm{regular} \mathrm{polygon} \mathrm{inscribed} \mathrm{in} \mathrm{a} \mathrm{circle} \mathrm{of} \mathrm{radius} R=\frac{n{R}^2}{2}\sin \frac{2\mathrm{\pi }}{n} \\ \mathrm{Area} \mathrm{of} \mathrm{a} \mathrm{regular} \mathrm{polygon} \mathrm{circumscribed} \mathrm{about} \mathrm{a} \mathrm{circle} \mathrm{of} \mathrm{radius} \mathrm{R}={\mathrm{nR}}^2\tan \frac{\mathrm{\pi }}{\mathrm{n}} \\ \frac{3}{4}=\frac{\frac{{\mathrm{nR}}^2}{2}\sin \frac{2\mathrm{\pi }}{\mathrm{n}}}{{\mathrm{nR}}^2\tan \frac{\mathrm{\pi }}{\mathrm{n}}}\Rightarrow \frac{3}{4}=\frac{1}{2\tan {\displaystyle \frac{\mathrm{\pi }}{\mathrm{n}}}}\left(\frac{2\tan \frac{\mathrm{\pi }}{\mathrm{n}}}{1+{\tan }^2\frac{\mathrm{\pi }}{\mathrm{n}}}\right) \\ \frac{3}{4}=\frac{1}{{\sec }^2{\displaystyle \frac{\mathrm{\pi }}{\mathrm{n}}}}\Rightarrow {\cos }^2\frac{\mathrm{\pi }}{\mathrm{n}}={\left(\frac{\sqrt{3}}{2}\right)}^2 \\ \therefore \mathrm{n}=6 \end{gathered}$$