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Question-208782

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Question Number 208782 by cherokeesay last updated on 22/Jun/24
Answered by mr W last updated on 23/Jun/24
Commented by cherokeesay last updated on 23/Jun/24
thank you master !
$${thank}\:{you}\:{master}\:! \\ $$
Commented by Tawa11 last updated on 23/Jun/24
Weldone sir.
$$\mathrm{Weldone}\:\mathrm{sir}. \\ $$
Commented by mr W last updated on 23/Jun/24
(r/R)=(√((25)/(49)))=(5/7) ⇒R=((7r)/5) d=2r (R−r)^2 +((√(15)))^2 =(3r−R)^2 25=4r^2 ⇒r=(5/2) ⇒R=(7/2) (√(R^2 −b^2 ))+(√(r^2 −b^2 ))=3r−R (√(R^2 −b^2 ))=4−(√(r^2 −b^2 )) 5=4(√(r^2 −b^2 )) b^2 =r^2 −((25)/(16)) ⇒b=((5(√3))/4) sin α_1 =(b/R)=((5(√3))/(14)) sin α_2 =(b/r)=((√3)/2) ⇒α_2 =(π/3) green area: (R^2 /4)(2α_1 −sin 2α_1 )+(r^2 /4)(2α_2 −sin 2α_2 ) =(7^2 /(4×4))(2sin^(−1) ((5(√3))/(14))−((2×5(√3)×11)/(14^2 )))+(5^2 /(4×4))(((2π)/3)−((√3)/2)) =((49)/8) sin^(−1) ((5(√3))/(14))+((25π)/(24))−((5(√3))/2)≈3.027
$$\frac{{r}}{{R}}=\sqrt{\frac{\mathrm{25}}{\mathrm{49}}}=\frac{\mathrm{5}}{\mathrm{7}}\:\Rightarrow{R}=\frac{\mathrm{7}{r}}{\mathrm{5}} \\ $$$${d}=\mathrm{2}{r} \\ $$$$\left({R}−{r}\right)^{\mathrm{2}} +\left(\sqrt{\mathrm{15}}\right)^{\mathrm{2}} =\left(\mathrm{3}{r}−{R}\right)^{\mathrm{2}} \\ $$$$\mathrm{25}=\mathrm{4}{r}^{\mathrm{2}} \: \\ $$$$\Rightarrow{r}=\frac{\mathrm{5}}{\mathrm{2}}\:\Rightarrow{R}=\frac{\mathrm{7}}{\mathrm{2}} \\ $$$$\sqrt{{R}^{\mathrm{2}} −{b}^{\mathrm{2}} }+\sqrt{{r}^{\mathrm{2}} −{b}^{\mathrm{2}} }=\mathrm{3}{r}−{R} \\ $$$$\sqrt{{R}^{\mathrm{2}} −{b}^{\mathrm{2}} }=\mathrm{4}−\sqrt{{r}^{\mathrm{2}} −{b}^{\mathrm{2}} } \\ $$$$\mathrm{5}=\mathrm{4}\sqrt{{r}^{\mathrm{2}} −{b}^{\mathrm{2}} } \\ $$$${b}^{\mathrm{2}} ={r}^{\mathrm{2}} −\frac{\mathrm{25}}{\mathrm{16}}\:\Rightarrow{b}=\frac{\mathrm{5}\sqrt{\mathrm{3}}}{\mathrm{4}} \\ $$$$\mathrm{sin}\:\alpha_{\mathrm{1}} =\frac{{b}}{{R}}=\frac{\mathrm{5}\sqrt{\mathrm{3}}}{\mathrm{14}} \\ $$$$\mathrm{sin}\:\alpha_{\mathrm{2}} =\frac{{b}}{{r}}=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:\Rightarrow\alpha_{\mathrm{2}} =\frac{\pi}{\mathrm{3}} \\ $$$${green}\:{area}: \\ $$$$\frac{{R}^{\mathrm{2}} }{\mathrm{4}}\left(\mathrm{2}\alpha_{\mathrm{1}} −\mathrm{sin}\:\mathrm{2}\alpha_{\mathrm{1}} \right)+\frac{{r}^{\mathrm{2}} }{\mathrm{4}}\left(\mathrm{2}\alpha_{\mathrm{2}} −\mathrm{sin}\:\mathrm{2}\alpha_{\mathrm{2}} \right) \\ $$$$=\frac{\mathrm{7}^{\mathrm{2}} }{\mathrm{4}×\mathrm{4}}\left(\mathrm{2sin}^{−\mathrm{1}} \frac{\mathrm{5}\sqrt{\mathrm{3}}}{\mathrm{14}}−\frac{\mathrm{2}×\mathrm{5}\sqrt{\mathrm{3}}×\mathrm{11}}{\mathrm{14}^{\mathrm{2}} }\right)+\frac{\mathrm{5}^{\mathrm{2}} }{\mathrm{4}×\mathrm{4}}\left(\frac{\mathrm{2}\pi}{\mathrm{3}}−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right) \\ $$$$=\frac{\mathrm{49}}{\mathrm{8}}\:\mathrm{sin}^{−\mathrm{1}} \frac{\mathrm{5}\sqrt{\mathrm{3}}}{\mathrm{14}}+\frac{\mathrm{25}\pi}{\mathrm{24}}−\frac{\mathrm{5}\sqrt{\mathrm{3}}}{\mathrm{2}}\approx\mathrm{3}.\mathrm{027} \\ $$

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