Question Number 179131 by cortano1 last updated on 25/Oct/22
Commented by cortano1 last updated on 25/Oct/22
$$\mathrm{find}\:\mathrm{the}\:\mathrm{area}\:\mathrm{of}\:\mathrm{circle} \\ $$
Commented by mr W last updated on 25/Oct/22
$${circle}\:{is}\:{not}\:{uniquely}\:{defined}. \\ $$
Commented by cortano1 last updated on 25/Oct/22
$$\mathrm{why}\:\mathrm{sir}? \\ $$
Commented by mr W last updated on 25/Oct/22
$${example}:\:{the}\:{green}\:{circle}\:{is}\:{also}\:{ok}. \\ $$
Commented by mr W last updated on 25/Oct/22
Commented by cortano1 last updated on 25/Oct/22
$$\mathrm{yes}\:\mathrm{sir}\:\mathrm{Mr}\:\mathrm{W}.\:\mathrm{Agree}\:\mathrm{with}\:\mathrm{you} \\ $$
Commented by mr W last updated on 25/Oct/22
$${with}\:{the}\:{given}\:{condition}\:{we}\:{have} \\ $$$${following}\:{largest}\:{circle}\:{and}\:{smallest} \\ $$$${circle}: \\ $$
Commented by mr W last updated on 25/Oct/22
Answered by a.lgnaoui last updated on 25/Oct/22
![∡COD=α CD=4 OC=OD=R CD^2 =(R−Rcos α)^2 +Rsin^2 α=2R^2 (1−cos α) cos α=1−(8/R^2 ) (1) BC^2 =R^2 (1−cos α)^2 +BN^2 ⇒ BN^2 =9−R^2 (1−cos α)^2 dautre part =BC^2 =OB^2 +OC^2 −2OBOCsin α OB=ON−BN=Rsin α−(√(9−R^2 (1−cos α)^2 )) OB^2 =R^2 sin^2 α+(9−R^2 (1−cos α)^2 −2Rsinα(√(9−R^2 (1−cos α)^2 )) =2R^2 cos α(1−cos α)+9−2Rsin α(√(9−R^2 (1−cosα)^2 )) BC^2 =R^2 +[2R^2 cos α(1−cos α)−2Rsin α(√(9−R^2 (1−cos α)^2 )) ] −2Rsin α[Rsin α−(√(9−R^2 (1−cos α)^2 )) ] BC^2 = R^2 +2R^2 cos α(1−cos α)−2R^2 sin^2 α+9 =R^2 +2R^2 cos α−2R^2 cos^2 α−2R^2 sin^2 α+9 9=2R^2 cos α−R^2 +9 cos α=(1/2)⇒ α=(π/3) (1)⇔(1/2)=1−(8/R^2 )⇒R=4 Donc aire du cercle =16π](https://www.tinkutara.com/question/Q179161.png)
$$\measuredangle\mathrm{COD}=\alpha\:\:\:\mathrm{CD}=\mathrm{4}\:\:\:\:\mathrm{OC}=\mathrm{OD}=\mathrm{R} \\ $$$$\mathrm{CD}^{\mathrm{2}} =\left(\mathrm{R}−\mathrm{Rcos}\:\alpha\right)^{\mathrm{2}} +\mathrm{Rsin}\:^{\mathrm{2}} \alpha=\mathrm{2R}^{\mathrm{2}} \left(\mathrm{1}−\mathrm{cos}\:\alpha\right) \\ $$$$\mathrm{cos}\:\alpha=\mathrm{1}−\frac{\mathrm{8}}{\mathrm{R}^{\mathrm{2}} }\:\:\:\:\:\:\:\:\:\:\left(\mathrm{1}\right) \\ $$$$\mathrm{BC}^{\mathrm{2}} =\mathrm{R}^{\mathrm{2}} \left(\mathrm{1}−\mathrm{cos}\:\alpha\right)^{\mathrm{2}} +\mathrm{BN}^{\mathrm{2}} \Rightarrow\:\:\:\mathrm{BN}^{\mathrm{2}} =\mathrm{9}−\mathrm{R}^{\mathrm{2}} \left(\mathrm{1}−\mathrm{cos}\:\alpha\right)^{\mathrm{2}} \\ $$$${dautre}\:{part} \\ $$$$=\mathrm{BC}^{\mathrm{2}} =\mathrm{OB}^{\mathrm{2}} +\mathrm{OC}^{\mathrm{2}} −\mathrm{2OBOCsin}\:\alpha \\ $$$$\mathrm{OB}=\mathrm{ON}−\mathrm{BN}=\mathrm{Rsin}\:\alpha−\sqrt{\mathrm{9}−\mathrm{R}^{\mathrm{2}} \left(\mathrm{1}−\mathrm{cos}\:\alpha\right)^{\mathrm{2}} }\: \\ $$$$\mathrm{OB}^{\mathrm{2}} =\mathrm{R}^{\mathrm{2}} \mathrm{sin}\:^{\mathrm{2}} \alpha+\left(\mathrm{9}−\mathrm{R}^{\mathrm{2}} \left(\mathrm{1}−\mathrm{cos}\:\alpha\right)^{\mathrm{2}} −\mathrm{2Rsin}\alpha\sqrt{\mathrm{9}−\mathrm{R}^{\mathrm{2}} \left(\mathrm{1}−\mathrm{cos}\:\alpha\right)^{\mathrm{2}} \:}\right. \\ $$$$ \\ $$$$=\mathrm{2R}^{\mathrm{2}} \mathrm{cos}\:\alpha\left(\mathrm{1}−\mathrm{cos}\:\alpha\right)+\mathrm{9}−\mathrm{2Rsin}\:\alpha\sqrt{\mathrm{9}−\mathrm{R}^{\mathrm{2}} \left(\mathrm{1}−\mathrm{cos}\alpha\right)^{\mathrm{2}} }\: \\ $$$$\mathrm{BC}^{\mathrm{2}} =\mathrm{R}^{\mathrm{2}} +\left[\mathrm{2R}^{\mathrm{2}} \mathrm{cos}\:\alpha\left(\mathrm{1}−\mathrm{cos}\:\alpha\right)−\mathrm{2Rsin}\:\alpha\sqrt{\mathrm{9}−\mathrm{R}^{\mathrm{2}} \left(\mathrm{1}−\mathrm{cos}\:\alpha\right)^{\mathrm{2}} }\:\right]\:\:\:−\mathrm{2Rsin}\:\alpha\left[\mathrm{Rsin}\:\alpha−\sqrt{\mathrm{9}−\mathrm{R}^{\mathrm{2}} \left(\mathrm{1}−\mathrm{cos}\:\alpha\right)^{\mathrm{2}} }\:\right] \\ $$$$\:\mathrm{BC}^{\mathrm{2}} =\:\:\mathrm{R}^{\mathrm{2}} +\mathrm{2R}^{\mathrm{2}} \mathrm{cos}\:\alpha\left(\mathrm{1}−\mathrm{cos}\:\alpha\right)−\mathrm{2R}^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} \alpha+\mathrm{9} \\ $$$$\:\:\:\:\:\:\:=\mathrm{R}^{\mathrm{2}} +\mathrm{2R}^{\mathrm{2}} \mathrm{cos}\:\alpha−\mathrm{2R}^{\mathrm{2}} \mathrm{cos}\:^{\mathrm{2}} \alpha−\mathrm{2R}^{\mathrm{2}} \mathrm{sin}\:^{\mathrm{2}} \alpha+\mathrm{9} \\ $$$$\mathrm{9}=\mathrm{2R}^{\mathrm{2}} \mathrm{cos}\:\alpha−\mathrm{R}^{\mathrm{2}} +\mathrm{9} \\ $$$$\mathrm{cos}\:\alpha=\frac{\mathrm{1}}{\mathrm{2}}\Rightarrow\:\:\:\alpha=\frac{\pi}{\mathrm{3}} \\ $$$$\:\left(\mathrm{1}\right)\Leftrightarrow\frac{\mathrm{1}}{\mathrm{2}}=\mathrm{1}−\frac{\mathrm{8}}{\mathrm{R}^{\mathrm{2}} }\Rightarrow\mathrm{R}=\mathrm{4} \\ $$$$\mathrm{Donc}\:\:\mathrm{aire}\:\mathrm{du}\:\mathrm{cercle}\:=\mathrm{16}\pi \\ $$$$ \\ $$
Commented by a.lgnaoui last updated on 25/Oct/22
Commented by mr W last updated on 25/Oct/22
$${as}\:{i}\:{said},\:{there}\:{is}\:{no}\:{unique}\:{solution}! \\ $$$$\mathrm{2}\leqslant{R}\leqslant\frac{\mathrm{3}+\sqrt{\mathrm{41}}}{\mathrm{2}}\:{is}\:{possible}. \\ $$
Commented by a.lgnaoui last updated on 25/Oct/22
![yes because the raport(3/4) can be presented with differentes Radius of curcles like the interals [2,((3+(√(41)))/2)] (en particulier R=4)](https://www.tinkutara.com/question/Q179171.png)
$${yes}\:\:{because}\:\:{the}\:{raport}\frac{\mathrm{3}}{\mathrm{4}}\:{can}\:{be}\: \\ $$$${presented}\:{with}\:{differentes}\:{Radius}\:{of}\:{curcles} \\ $$$${like}\:{the}\:{interals}\:\left[\mathrm{2},\frac{\mathrm{3}+\sqrt{\mathrm{41}}}{\mathrm{2}}\right] \\ $$$$\left({en}\:{particulier}\:{R}=\mathrm{4}\right) \\ $$$$ \\ $$
Commented by mr W last updated on 25/Oct/22
$${yes}. \\ $$