Muntazam sakkizburchakka ichki va tashqi aylanalar chizilgan. Agar ichki chizilgan aylananing radiusi r=2+√2…

Halqa yuzi $\pi(R^2-r^2)=\pi\left(\frac a2\right)^2$, bunda $a$ — sakkizburchak tomoni. $r=\frac a2\,\mathrm{ctg}\,22{,}5^\circ=\frac a2(\sqrt2+1)$, bundan $\frac a2=\frac{2+\sqrt2}{\sqrt2+1}=\sqrt2$. Yuza $2\pi$.

Manba: UZBMB namunaviy test topshiriqlari (Milliy sertifikat, Matematika)