A regular pentagon and a regular decagon have the same perimeter, tehn their areas are in the ratio $=\frac{K}{\sqrt{5}}$, then $K=$ _____

Number of sides in pentagon $\left(n\right)=5$

Pentagon perimeter $=$ decagon perimeter

$5a_D=10a_D$

$a_P=2a_D$

$\frac{areaofpentagon}{areaofdeeagon}=\frac{5{\left(\frac{2a_D}{4}\right)}^2\cot \frac{\pi }{5}}{10{\left(\frac{aD}{4}\right)}^2.\cot \frac{\pi }{10}}$

$=\frac{5\left(\frac{4a_D^2}{16}\right).\cot \frac{\pi }{5}}{10\left(\frac{a_D^2}{16}\right).\cot \frac{\pi }{10}}$

$=\frac{20\left(a_D^2\right)\cot \left(\frac{\pi }{5}\right)}{10\left(\frac{a_D^2}{16}\right)\cot \left(\frac{\pi }{10}\right)}$

$=\frac{2\cot 36°}{\tan 18°}\left[\therefore \frac{{\displaystyle \pi }}{{\displaystyle 5}}=36°and\frac{{\displaystyle \pi }}{{\displaystyle 10}}=18°\right]$

$=\frac{2\tan 18°}{\tan 36°}\left[\therefore \tan 18°=\frac{{\displaystyle \sin 18°}}{{\displaystyle \cos 18°}}\&\tan 36°=\frac{{\displaystyle \sin 36°}}{{\displaystyle \cos 36°}}\right]$

$\begin{array}{l}=\frac{2(\sqrt{5}-1)}{\sqrt{10+2\sqrt{5}}}\times \frac{\sqrt{5}+1}{\sqrt{10-2\sqrt{5}}}\\ =\frac{2(5-1)}{\sqrt{100-20}}=\frac{2\left(4\right)}{\sqrt{80}}\\ =\frac{8}{4\sqrt{5}}=\frac{2}{\sqrt{5}}=\frac{K}{\sqrt{5}}\end{array}$

Then $K=2$