Question Number 187284 by Rupesh123 last updated on 15/Feb/23
Commented by mr W last updated on 15/Feb/23
$${i}\:{think}\:{the}\:{green}\:{area}\:{is}\:{not}\:{unique}. \\ $$
Commented by mr W last updated on 16/Feb/23
$${R}={radius} \\ $$$$\mathrm{sin}\:\theta=\frac{\mathrm{12}}{\mathrm{2}{R}}=\frac{\mathrm{6}}{{R}} \\ $$$${A}_{{green}} =\frac{{R}\left(\mathrm{2}{R}\:\mathrm{cos}\:\theta−\mathrm{9}\right)\:\mathrm{sin}\:\theta}{\mathrm{2}} \\ $$$$\:\:\:\:=\frac{{R}\left(\mathrm{2}{R}\:\frac{\sqrt{{R}^{\mathrm{2}} −\mathrm{6}^{\mathrm{2}} }}{{R}}−\mathrm{9}\right)×\frac{\mathrm{6}}{{R}}}{\mathrm{2}} \\ $$$$\:\:\:\:=\mathrm{6}\sqrt{{R}^{\mathrm{2}} −\mathrm{6}^{\mathrm{2}} }−\mathrm{27} \\ $$$${we}\:{see}\:{the}\:{green}\:{area}\:{is}\:{not}\:{unique}. \\ $$$${for}\:{any}\:{circle}\:{with} \\ $$$${R}>\sqrt{\mathrm{6}^{\mathrm{2}} +\left(\frac{\mathrm{9}}{\mathrm{2}}\right)^{\mathrm{2}} }=\frac{\mathrm{15}}{\mathrm{2}}=\mathrm{7}.\mathrm{5} \\ $$$${we}\:{can}\:{get}\:{a}\:{different}\:{green}\:{area}. \\ $$
Answered by a.lgnaoui last updated on 16/Feb/23
$$\bigtriangleup{ABC}\:\:\bigtriangleup{OIC}\:{semblables} \\ $$$$\frac{{AB}}{\mathrm{2}{R}}=\frac{{OI}}{{R}}\Rightarrow\:\:\:{OI}=\frac{{AB}}{\mathrm{2}}\:\:=\mathrm{6}\:\:\:\:\:\:\: \\ $$$${Area}\left({OIC}\right)=\frac{{CD}×{OI}}{\mathrm{2}}=\mathrm{3}{CD}\:\:\left({i}\right) \\ $$$${BD}^{\mathrm{2}} =\mathrm{9}^{\mathrm{2}} +\mathrm{12}^{\mathrm{2}} \:\:\:\:\Rightarrow{BD}=\mathrm{15} \\ $$$$\bigtriangleup{BDC}\:\:\: \\ $$$$\:{BC}^{\mathrm{2}} =\mathrm{15}^{\mathrm{2}} +{CD}^{\mathrm{2}} −\mathrm{30}{CD}\mathrm{cos}\:\lambda\:\:\left(\mathrm{1}\right) \\ $$$$\bigtriangleup{ABC}\: \\ $$$$\:{BC}^{\mathrm{2}} =\mathrm{12}^{\mathrm{2}} +\left(\mathrm{9}+{CD}\right)^{\mathrm{2}} \:\:\:\:\:\:\left(\mathrm{2}\right) \\ $$$$\bigtriangleup{BDC}\:\:\frac{\mathrm{sin}\:\alpha}{{BD}}=\frac{\mathrm{sin}\:\lambda}{{BC}}\:=\frac{\mathrm{sin}\:\beta}{{CD}}\:\Rightarrow\mathrm{sin}\:\alpha=\frac{\mathrm{12sin}\:\lambda}{{BC}} \\ $$$$\frac{\mathrm{sin}\:\beta}{{CD}}=\frac{\mathrm{sin}\:\lambda}{{BC}}\Rightarrow{CD}\mathrm{sin}\:\lambda={BC}\mathrm{sin}\:\beta \\ $$$${BC}=\frac{{CD}\mathrm{sin}\:\lambda}{\mathrm{sin}\:\beta}\Rightarrow{CD}=\frac{{BC}}{\mathrm{tan}\:\lambda}\:=\frac{{BD}\mathrm{sin}\:\beta}{\mathrm{sin}\:\alpha} \\ $$$$\bigtriangleup{ABD}\:\:\mathrm{tan}\:\:\varphi=\frac{\mathrm{4}}{\mathrm{3}}\:\:\varphi=\mathrm{53},\mathrm{13} \\ $$$$\Rightarrow\lambda=\mathrm{126},\mathrm{87} \\ $$$$\Rightarrow\lambda=\mathrm{180}−\mathrm{53},\mathrm{13}=\mathrm{126},\mathrm{87} \\ $$$$\:\Rightarrow\:\mathrm{tan}\:\beta=\frac{\mathrm{9}}{\mathrm{12}}=\frac{\mathrm{3}}{\mathrm{4}}\:\:\:\beta=\mathrm{36},\mathrm{87} \\ $$$$\alpha+\beta+\lambda=\mathrm{180}\:\:\alpha=\mathrm{16},\mathrm{25} \\ $$$$\mathrm{180}−\mathrm{163},\mathrm{87}=\mathrm{16},\mathrm{25} \\ $$$$\mathrm{sin}\:\alpha=\frac{\mathrm{6}}{{R}}\:\:\:\:\:{R}=\frac{\mathrm{6}}{\mathrm{sin}\:\alpha}= \\ $$$${R}=\frac{\mathrm{6}}{\mathrm{0},\mathrm{28}}=\frac{\mathrm{150}}{\mathrm{7}}=\mathrm{22},\mathrm{125} \\ $$$$\mathrm{4}{R}^{\mathrm{2}} =\mathrm{225}+{CD}^{\mathrm{2}} +\mathrm{18}{CD} \\ $$$$\mathrm{1835}={x}^{\mathrm{2}} +\mathrm{18}{x}+\mathrm{225} \\ $$$${x}^{\mathrm{2}} +\mathrm{18}{x}−\mathrm{1610} \\ $$$${x}=−\mathrm{9}\pm\mathrm{41},\mathrm{12}=\mathrm{32},\mathrm{12} \\ $$$$\Rightarrow\:\:\:\:\:{CD}=\mathrm{32},\mathrm{12} \\ $$$${Area}=\mathrm{3}×\mathrm{32}=\mathrm{96},\mathrm{36}\:\:\:? \\ $$$$ \\ $$
Commented by a.lgnaoui last updated on 16/Feb/23
Commented by a.lgnaoui last updated on 16/Feb/23
Commented by mr W last updated on 16/Feb/23
$${wrong}! \\ $$$${it}\:{is}\:{proved}\:{that}\:{no}\:{unique}\:{solution} \\ $$$${exists}.\:{just}\:{try}\:{with}\:{R}=\mathrm{7}.\mathrm{5}\:{and}\:{then} \\ $$$${with}\:{R}=\mathrm{10}\:{and}\:{then}\:{with}\:{R}=\mathrm{20}! \\ $$
Commented by a.lgnaoui last updated on 16/Feb/23
$${yes}\: \\ $$