Question Number 90594 by byaw last updated on 24/Apr/20
Commented by mr W last updated on 24/Apr/20
Commented by mr W last updated on 24/Apr/20
$${when}\:{posting}\:{an}\:{image},\:{you}\:{can}\: \\ $$$${rotate}\:{it}\:{at}\:{first}\:{as}\:{shown}\:{above}. \\ $$
Commented by mahdi last updated on 25/Apr/20
$$\mathrm{AB}=\mathrm{2r}=\mathrm{x}+\mathrm{1}\Rightarrow\mathrm{r}=\frac{\mathrm{x}+\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{CD}=\sqrt{\mathrm{OD}^{\mathrm{2}} −\mathrm{OC}^{\mathrm{2}} }=\sqrt{\mathrm{r}^{\mathrm{2}} −\left(\mathrm{x}−\mathrm{r}\right)^{\mathrm{2}} } \\ $$$$=\sqrt{\mathrm{2rx}−\mathrm{x}^{\mathrm{2}} }=\sqrt{\mathrm{2}\left(\frac{\mathrm{x}+\mathrm{1}}{\mathrm{2}}\right)\mathrm{x}−\mathrm{x}^{\mathrm{2}} }=\sqrt{\mathrm{x}} \\ $$