Question Number 194116 by universe last updated on 28/Jun/23
Answered by mr W last updated on 02/Jul/23
Commented by mr W last updated on 02/Jul/23
$${R}={radius}\:{of}\:{semicircle} \\ $$$${r}={radius}\:{of}\:{small}\:{circle} \\ $$$${AD}=\mathrm{2}{R} \\ $$$${OF}={R}−{r} \\ $$$${OE}=\sqrt{\left({R}−{r}\right)^{\mathrm{2}} −{r}^{\mathrm{2}} } \\ $$$${AB}={R}−{r}+\sqrt{\left({R}−{r}\right)^{\mathrm{2}} −{r}^{\mathrm{2}} } \\ $$$${CD}={R}−{r}−\sqrt{\left({R}−{r}\right)^{\mathrm{2}} −{r}^{\mathrm{2}} } \\ $$$${AB}×{CD}={r}^{\mathrm{2}} \\ $$$${AB}+{CD}+\mathrm{2}\sqrt{{AB}×{CD}}=\mathrm{2}\left({R}−{r}\right)+\mathrm{2}{r}=\mathrm{2}{R}={AD} \\ $$$${i}.{e}.\:\left(\sqrt{{AB}}+\sqrt{{CD}}\right)^{\mathrm{2}} =\left(\sqrt{{AD}}\right)^{\mathrm{2}} \\ $$$$\Rightarrow\sqrt{{AD}}=\sqrt{{AB}}+\sqrt{{CD}} \\ $$