\[Схематический\ рисунок.\]
\[Дано:\]
\[AB - сторона;\]
\[OH\bot AB;\]
\[OH = r;\]
\[AB = a;\]
\[OA = OB = R.\]
\[Найти:\]
\[\text{R.}\]
\[Решение.\]
\[1)\ \mathrm{\Delta}AOB - равнобедренный:\]
\[AO = OB = R.\]
\[Отсюда:\]
\[AH = BH = \frac{1}{2}AB = \frac{1}{2}\text{a.}\]
\[2)\ OB = \sqrt{OH^{2} + BH^{2}}\]
\[OB = R = \sqrt{r^{2} + \frac{a^{2}}{4}}.\]
\[Ответ:\ \ \sqrt{r^{2} + \frac{a^{2}}{4}}.\]