Question Number 156913 by cortano last updated on 17/Oct/21
Answered by mr W last updated on 17/Oct/21
Commented by cortano last updated on 17/Oct/21
$${blue}\:{length}\:=\:\mathrm{10}\:.\:{yes}\:{thanks} \\ $$
Commented by mr W last updated on 17/Oct/21
![OD=(√((R−r)^2 −r^2 ))=(√(R(R−2r))) FD=R+(√(R(R−2r)))=blue OE=OD−r=(√(R(R−2r)))−r EG=R−OE FE=R+OE AE^2 =FE×EG=R^2 −OE^2 FA^2 =AE^2 +FE^2 =R^2 −OE^2 +FE^2 FA^2 =R^2 −OE^2 +(R+OE)^2 FA^2 =2R^2 +2R×OE FA^2 =2R^2 +2R×[(√(R(R−2r)))−r] FA^2 =R^2 +2R(√(R(R−2r)))+R(R−2r) FA^2 =[R+(√(R(R−2r)))]^2 FA=R+(√(R(R−2r)))=gree it is to see FD=FA, i.e. blue=green.](https://www.tinkutara.com/question/Q156920.png)
$${OD}=\sqrt{\left({R}−{r}\right)^{\mathrm{2}} −{r}^{\mathrm{2}} }=\sqrt{{R}\left({R}−\mathrm{2}{r}\right)} \\ $$$${FD}={R}+\sqrt{{R}\left({R}−\mathrm{2}{r}\right)}={blue} \\ $$$${OE}={OD}−{r}=\sqrt{{R}\left({R}−\mathrm{2}{r}\right)}−{r} \\ $$$${EG}={R}−{OE} \\ $$$${FE}={R}+{OE} \\ $$$${AE}^{\mathrm{2}} ={FE}×{EG}={R}^{\mathrm{2}} −{OE}^{\mathrm{2}} \\ $$$${FA}^{\mathrm{2}} ={AE}^{\mathrm{2}} +{FE}^{\mathrm{2}} ={R}^{\mathrm{2}} −{OE}^{\mathrm{2}} +{FE}^{\mathrm{2}} \\ $$$${FA}^{\mathrm{2}} ={R}^{\mathrm{2}} −{OE}^{\mathrm{2}} +\left({R}+{OE}\right)^{\mathrm{2}} \\ $$$${FA}^{\mathrm{2}} =\mathrm{2}{R}^{\mathrm{2}} +\mathrm{2}{R}×{OE} \\ $$$${FA}^{\mathrm{2}} =\mathrm{2}{R}^{\mathrm{2}} +\mathrm{2}{R}×\left[\sqrt{{R}\left({R}−\mathrm{2}{r}\right)}−{r}\right] \\ $$$${FA}^{\mathrm{2}} ={R}^{\mathrm{2}} +\mathrm{2}{R}\sqrt{{R}\left({R}−\mathrm{2}{r}\right)}+{R}\left({R}−\mathrm{2}{r}\right) \\ $$$${FA}^{\mathrm{2}} =\left[{R}+\sqrt{{R}\left({R}−\mathrm{2}{r}\right)}\right]^{\mathrm{2}} \\ $$$${FA}={R}+\sqrt{{R}\left({R}−\mathrm{2}{r}\right)}={gree} \\ $$$${it}\:{is}\:{to}\:{see}\:{FD}={FA},\:{i}.{e}.\:{blue}={green}. \\ $$
Commented by Tawa11 last updated on 17/Oct/21
$$\mathrm{great}\:\mathrm{sir} \\ $$