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Question Number 187284 by Rupesh123 last updated on 15/Feb/23

Commented by mr W last updated on 15/Feb/23

$$ithinkthegreenareaisnotunique.$$

Commented by mr W last updated on 16/Feb/23

$$R=radius$$$$\sin \theta =\frac{12}{2R}=\frac{6}{R}$$$$A_{green}=\frac{R(2R\cos \theta -9)\sin \theta }{2}$$$$=\frac{R(2R\frac{\sqrt{R^2-6^2}}{R}-9)\times \frac{6}{R}}{2}$$$$=6\sqrt{R^2-6^2}-27$$$$weseethegreenareaisnotunique.$$$$foranycirclewith$$$$R>\sqrt{6^2+{\left(\frac{9}{2}\right)}^2}=\frac{15}{2}=7.5$$$$wecangetadifferentgreenarea.$$

Answered by a.lgnaoui last updated on 16/Feb/23

$$△ABC△OICsemblables$$$$\frac{AB}{2R}=\frac{OI}{R}\Rightarrow OI=\frac{AB}{2}=6$$$$Area\left(OIC\right)=\frac{CD\times OI}{2}=3CD\left(i\right)$$$${BD}^2=9^2+{12}^2\Rightarrow BD=15$$$$△BDC$$$${BC}^2={15}^2+{CD}^2-30CD\cos \lambda \left(1\right)$$$$△ABC$$$${BC}^2={12}^2+{(9+CD)}^2\left(2\right)$$$$△BDC\frac{\sin \alpha }{BD}=\frac{\sin \lambda }{BC}=\frac{\sin \beta }{CD}\Rightarrow \sin \alpha =\frac{12\sin \lambda }{BC}$$$$\frac{\sin \beta }{CD}=\frac{\sin \lambda }{BC}\Rightarrow CD\sin \lambda =BC\sin \beta$$$$BC=\frac{CD\sin \lambda }{\sin \beta }\Rightarrow CD=\frac{BC}{\tan \lambda }=\frac{BD\sin \beta }{\sin \alpha }$$$$△ABD\tan \phi =\frac{4}{3}\phi =53,13$$$$\Rightarrow \lambda =126,87$$$$\Rightarrow \lambda =180-53,13=126,87$$$$\Rightarrow \tan \beta =\frac{9}{12}=\frac{3}{4}\beta =36,87$$$$\alpha +\beta +\lambda =180\alpha =16,25$$$$180-163,87=16,25$$$$\sin \alpha =\frac{6}{R}R=\frac{6}{\sin \alpha }=$$$$R=\frac{6}{0,28}=\frac{150}{7}=22,125$$$$4R^2=225+{CD}^2+18CD$$$$1835=x^2+18x+225$$$$x^2+18x-1610$$$$x=-9\pm 41,12=32,12$$$$\Rightarrow CD=32,12$$$$Area=3\times 32=96,36?$$$$$$

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

$$wrong!$$$$itisprovedthatnouniquesolution$$$$exists.justtrywithR=7.5andthen$$$$withR=10andthenwithR=20!$$

Commented by a.lgnaoui last updated on 16/Feb/23

$$yes$$

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