Question Number 215195 by mr W last updated on 31/Dec/24
Commented by mr W last updated on 31/Dec/24
$${find}\:{the}\:{shaded}\:{area}=? \\ $$
Answered by mr W last updated on 02/Jan/25
Commented by mr W last updated on 31/Dec/24
Commented by mr W last updated on 02/Jan/25
![R=d+r+2r_1 =4+d λ=((R^2 +r^2 −d^2 )/(2Rr))=((5+2d)/(4+d)) cosh^(−1) λ=2 ln [tan ((π/4)+(π/(2×6)))]=ln 3 λ+(√(λ^2 −1))=3 λ^2 −1=(3−λ)^2 =9−6λ+λ^2 ⇒λ=(5/3)=((5+2d)/(4+d)) ⇒d=5 ⇒R=9 r_4 +r+r_1 =R ⇒r_4 =6 ((d^2 +(r+r_2 )^2 −(R−r_2 )^2 )/(2d(r+r_2 )))=−(((r+r_2 )^2 +(r+r_1 )^2 −(r_1 +r_2 )^2 )/(2(r+r_2 )(r+r_1 ))) ⇒r_2 =((24)/(19)) ((d^2 +(R−r_3 )^2 −(r+r_3 )^2 )/(2d(R−r_3 )))=−(((R−r_3 )^2 +(R−r_4 )^2 −(r_4 +r_3 )^2 )/(2(R−r_3 )(R−r_4 ))) ⇒r_3 =(8/3) cos γ=(((R−r_4 )^2 +(R−r_3 )^2 −(r_4 +r_3 )^2 )/(2(R−r_4 )(R−r_3 ))) =−((13)/(19)) ⇒sin γ=((8(√3))/(19)) ⇒γ=π−sin^(−1) ((8(√3))/(19)) sin α=(((R−r_3 )sin γ)/(r_4 +r_3 ))=(((9−(8/3)))/(6+(8/3)))×((8(√3))/(19))=((4(√3))/(13)) sin β=(((R−r_4 )sin γ)/(r_4 +r_3 ))=(((9−6))/(6+(8/3)))×((8(√3))/(19))=((36(√3))/(247)) A_(shade) =(1/2)[γR^2 −(π−α)r_4 ^2 −(π−β)r_3 ^2 −(R−r_4 )(R−r_3 )sin γ] =(1/2)[9^2 (π−sin^(−1) ((8(√3))/(19)))−6^2 (π−sin^(−1) ((4(√3))/(13)))−((8/3))^2 (π−sin^(−1) ((36(√3))/(247)))−(9−6)(9−(8/3))((8(√3))/(19))] =((341π)/(18))−4(√3)−((81)/2) sin^(−1) ((8(√3))/(19))+18 sin^(−1) ((4(√3))/(13))+((32)/9) sin^(−1) ((36(√3))/(247)) ≈30.512513235](https://www.tinkutara.com/question/Q215198.png)
$${R}={d}+{r}+\mathrm{2}{r}_{\mathrm{1}} =\mathrm{4}+{d} \\ $$$$\lambda=\frac{{R}^{\mathrm{2}} +{r}^{\mathrm{2}} −{d}^{\mathrm{2}} }{\mathrm{2}{Rr}}=\frac{\mathrm{5}+\mathrm{2}{d}}{\mathrm{4}+{d}} \\ $$$$\mathrm{cosh}^{−\mathrm{1}} \:\lambda=\mathrm{2}\:\mathrm{ln}\:\left[\mathrm{tan}\:\left(\frac{\pi}{\mathrm{4}}+\frac{\pi}{\mathrm{2}×\mathrm{6}}\right)\right]=\mathrm{ln}\:\mathrm{3} \\ $$$$\lambda+\sqrt{\lambda^{\mathrm{2}} −\mathrm{1}}=\mathrm{3} \\ $$$$\lambda^{\mathrm{2}} −\mathrm{1}=\left(\mathrm{3}−\lambda\right)^{\mathrm{2}} =\mathrm{9}−\mathrm{6}\lambda+\lambda^{\mathrm{2}} \\ $$$$\Rightarrow\lambda=\frac{\mathrm{5}}{\mathrm{3}}=\frac{\mathrm{5}+\mathrm{2}{d}}{\mathrm{4}+{d}} \\ $$$$\Rightarrow{d}=\mathrm{5} \\ $$$$\Rightarrow{R}=\mathrm{9} \\ $$$${r}_{\mathrm{4}} +{r}+{r}_{\mathrm{1}} ={R}\:\Rightarrow{r}_{\mathrm{4}} =\mathrm{6} \\ $$$$\frac{{d}^{\mathrm{2}} +\left({r}+{r}_{\mathrm{2}} \right)^{\mathrm{2}} −\left({R}−{r}_{\mathrm{2}} \right)^{\mathrm{2}} }{\mathrm{2}{d}\left({r}+{r}_{\mathrm{2}} \right)}=−\frac{\left({r}+{r}_{\mathrm{2}} \right)^{\mathrm{2}} +\left({r}+{r}_{\mathrm{1}} \right)^{\mathrm{2}} −\left({r}_{\mathrm{1}} +{r}_{\mathrm{2}} \right)^{\mathrm{2}} }{\mathrm{2}\left({r}+{r}_{\mathrm{2}} \right)\left({r}+{r}_{\mathrm{1}} \right)} \\ $$$$\Rightarrow{r}_{\mathrm{2}} =\frac{\mathrm{24}}{\mathrm{19}} \\ $$$$\frac{{d}^{\mathrm{2}} +\left({R}−{r}_{\mathrm{3}} \right)^{\mathrm{2}} −\left({r}+{r}_{\mathrm{3}} \right)^{\mathrm{2}} }{\mathrm{2}{d}\left({R}−{r}_{\mathrm{3}} \right)}=−\frac{\left({R}−{r}_{\mathrm{3}} \right)^{\mathrm{2}} +\left({R}−{r}_{\mathrm{4}} \right)^{\mathrm{2}} −\left({r}_{\mathrm{4}} +{r}_{\mathrm{3}} \right)^{\mathrm{2}} }{\mathrm{2}\left({R}−{r}_{\mathrm{3}} \right)\left({R}−{r}_{\mathrm{4}} \right)} \\ $$$$\Rightarrow{r}_{\mathrm{3}} =\frac{\mathrm{8}}{\mathrm{3}} \\ $$$$\mathrm{cos}\:\gamma=\frac{\left({R}−{r}_{\mathrm{4}} \right)^{\mathrm{2}} +\left({R}−{r}_{\mathrm{3}} \right)^{\mathrm{2}} −\left({r}_{\mathrm{4}} +{r}_{\mathrm{3}} \right)^{\mathrm{2}} }{\mathrm{2}\left({R}−{r}_{\mathrm{4}} \right)\left({R}−{r}_{\mathrm{3}} \right)} \\ $$$$\:\:\:\:\:\:\:\:\:\:=−\frac{\mathrm{13}}{\mathrm{19}}\: \\ $$$$\Rightarrow\mathrm{sin}\:\gamma=\frac{\mathrm{8}\sqrt{\mathrm{3}}}{\mathrm{19}}\:\Rightarrow\gamma=\pi−\mathrm{sin}^{−\mathrm{1}} \frac{\mathrm{8}\sqrt{\mathrm{3}}}{\mathrm{19}} \\ $$$$\mathrm{sin}\:\alpha=\frac{\left({R}−{r}_{\mathrm{3}} \right)\mathrm{sin}\:\gamma}{{r}_{\mathrm{4}} +{r}_{\mathrm{3}} }=\frac{\left(\mathrm{9}−\frac{\mathrm{8}}{\mathrm{3}}\right)}{\mathrm{6}+\frac{\mathrm{8}}{\mathrm{3}}}×\frac{\mathrm{8}\sqrt{\mathrm{3}}}{\mathrm{19}}=\frac{\mathrm{4}\sqrt{\mathrm{3}}}{\mathrm{13}} \\ $$$$\mathrm{sin}\:\beta=\frac{\left({R}−{r}_{\mathrm{4}} \right)\mathrm{sin}\:\gamma}{{r}_{\mathrm{4}} +{r}_{\mathrm{3}} }=\frac{\left(\mathrm{9}−\mathrm{6}\right)}{\mathrm{6}+\frac{\mathrm{8}}{\mathrm{3}}}×\frac{\mathrm{8}\sqrt{\mathrm{3}}}{\mathrm{19}}=\frac{\mathrm{36}\sqrt{\mathrm{3}}}{\mathrm{247}} \\ $$$${A}_{{shade}} =\frac{\mathrm{1}}{\mathrm{2}}\left[\gamma{R}^{\mathrm{2}} −\left(\pi−\alpha\right){r}_{\mathrm{4}} ^{\mathrm{2}} −\left(\pi−\beta\right){r}_{\mathrm{3}} ^{\mathrm{2}} −\left({R}−{r}_{\mathrm{4}} \right)\left({R}−{r}_{\mathrm{3}} \right)\mathrm{sin}\:\gamma\right] \\ $$$$\:\:\:=\frac{\mathrm{1}}{\mathrm{2}}\left[\mathrm{9}^{\mathrm{2}} \left(\pi−\mathrm{sin}^{−\mathrm{1}} \frac{\mathrm{8}\sqrt{\mathrm{3}}}{\mathrm{19}}\right)−\mathrm{6}^{\mathrm{2}} \left(\pi−\mathrm{sin}^{−\mathrm{1}} \frac{\mathrm{4}\sqrt{\mathrm{3}}}{\mathrm{13}}\right)−\left(\frac{\mathrm{8}}{\mathrm{3}}\right)^{\mathrm{2}} \left(\pi−\mathrm{sin}^{−\mathrm{1}} \frac{\mathrm{36}\sqrt{\mathrm{3}}}{\mathrm{247}}\right)−\left(\mathrm{9}−\mathrm{6}\right)\left(\mathrm{9}−\frac{\mathrm{8}}{\mathrm{3}}\right)\frac{\mathrm{8}\sqrt{\mathrm{3}}}{\mathrm{19}}\right] \\ $$$$\:\:\:=\frac{\mathrm{341}\pi}{\mathrm{18}}−\mathrm{4}\sqrt{\mathrm{3}}−\frac{\mathrm{81}}{\mathrm{2}}\:\mathrm{sin}^{−\mathrm{1}} \frac{\mathrm{8}\sqrt{\mathrm{3}}}{\mathrm{19}}+\mathrm{18}\:\mathrm{sin}^{−\mathrm{1}} \frac{\mathrm{4}\sqrt{\mathrm{3}}}{\mathrm{13}}+\frac{\mathrm{32}}{\mathrm{9}}\:\mathrm{sin}^{−\mathrm{1}} \frac{\mathrm{36}\sqrt{\mathrm{3}}}{\mathrm{247}} \\ $$$$\:\:\:\approx\mathrm{30}.\mathrm{512513235} \\ $$