Question Number 162473 by amin96 last updated on 29/Dec/21
Answered by mr W last updated on 29/Dec/21
Commented by amin96 last updated on 29/Dec/21
$$\boldsymbol{\mathrm{Thanks}}\:\boldsymbol{\mathrm{alot}}\:\boldsymbol{\mathrm{sir}}\:\boldsymbol{\mathrm{mr}}\:\boldsymbol{\mathrm{W}} \\ $$
Commented by mr W last updated on 29/Dec/21
$$\mathrm{cos}\:\alpha=\frac{\mathrm{2}}{\:\mathrm{6}}=\frac{\mathrm{1}}{\:\mathrm{3}} \\ $$$$\mathrm{tan}\:\frac{\alpha}{\mathrm{2}}=\sqrt{\frac{\mathrm{1}−\mathrm{cos}\:\alpha}{\mathrm{1}+\mathrm{cos}\:\alpha}}=\sqrt{\frac{\mathrm{3}−\mathrm{1}}{\:\mathrm{3}+\mathrm{1}}}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}} \\ $$$${a}={radius}\:{of}\:{green}\:{circles} \\ $$$${AB}=\frac{{a}}{\mathrm{tan}\:\frac{\alpha}{\mathrm{2}}}=\sqrt{\mathrm{2}}{a} \\ $$$${OA}=\frac{\mathrm{3}}{\mathrm{tan}\:\alpha}=\frac{\mathrm{3}}{\mathrm{2}\sqrt{\mathrm{2}}}=\frac{\mathrm{3}\sqrt{\mathrm{2}}}{\mathrm{4}} \\ $$$${OB}=\frac{\mathrm{3}\sqrt{\mathrm{2}}}{\mathrm{4}}+\sqrt{\mathrm{2}}{a} \\ $$$${OC}=\mathrm{3}−{a} \\ $$$$\left(\mathrm{3}−{a}\right)^{\mathrm{2}} ={a}^{\mathrm{2}} +\left(\frac{\mathrm{3}\sqrt{\mathrm{2}}}{\mathrm{4}}+\sqrt{\mathrm{2}}{a}\right)^{\mathrm{2}} \\ $$$$\mathrm{2}{a}^{\mathrm{2}} +\mathrm{9}{a}−\frac{\mathrm{63}}{\mathrm{8}}=\mathrm{0} \\ $$$${a}=\frac{−\mathrm{9}+\mathrm{12}}{\mathrm{4}}=\frac{\mathrm{3}}{\mathrm{4}} \\ $$$${A}_{{green}} =\mathrm{2}×\pi\left(\frac{\mathrm{3}}{\mathrm{4}}\right)^{\mathrm{2}} =\frac{\mathrm{9}\pi}{\mathrm{8}} \\ $$
Commented by Tawa11 last updated on 30/Dec/21
$$\mathrm{Great}\:\mathrm{sir} \\ $$