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Question-190094

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Question Number 190094 by mnjuly1970 last updated on 27/Mar/23
Answered by a.lgnaoui last updated on 28/Mar/23
△AEF EF^2 =AE^2 +AF^2 AE=R ;AF=2R−r (r=FB) ⇒(R+r)^2 =R^2 +(2R−r)^2 2Rr=4R^2 −4Rr r=(2/3)R FB∣∣ CD ∡CGD=∡FGB ((FB)/(CD))=((FG)/(CG))=((BG)/(GD)) GD=BD−BG=2R(√2) −3 FB=r CD=2R (r/(2R))=(3/(2R(√2) −3)) (1/3)=(3/(2R(√2) −3)) ⇒ R=3(√2) (i) △CDE tan α=(1/2) CE=R(√5) △DEH ((cos α)/(DH))=((sin 45)/(EH)) (1) △CDH ((sin α)/(DH))=((sin 45)/(R(√5) −EH)) (2) 1 EH=((DHsin 45)/(cos α)) DHsin 45=(R(√5) −EH)sin α EH=(1/(sin α))(R(√(5 )) sinα−DHsin 45) ⇒((DHsin 45)/(cos α))=((R(√5) sin α−DHsin 45)/(sin α)) tan α=((R(√5) sin α−DHsin 45)/(DHsin 45)) (1/2)=((R(√5) ×(R/(R(√5)))−((DH(√2))/2))/((DH(√2))/2))⇒DH=((4R)/(3(√2))) (i)R=3(√2) ⇒ DH=4
$$\bigtriangleup{AEF}\:\:\:{EF}^{\mathrm{2}} ={AE}^{\mathrm{2}} +{AF}^{\mathrm{2}} \\ $$$${AE}={R}\:;{AF}=\mathrm{2}{R}−{r}\:\:\:\left({r}={FB}\right) \\ $$$$\Rightarrow\left({R}+{r}\right)^{\mathrm{2}} ={R}^{\mathrm{2}} +\left(\mathrm{2}{R}−{r}\right)^{\mathrm{2}} \\ $$$$\:\:\:\:\:\mathrm{2}{Rr}=\mathrm{4}{R}^{\mathrm{2}} −\mathrm{4}{Rr}\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:{r}=\frac{\mathrm{2}}{\mathrm{3}}{R} \\ $$$${FB}\mid\mid\:{CD}\:\:\:\:\measuredangle{CGD}=\measuredangle{FGB} \\ $$$$\frac{{FB}}{{CD}}=\frac{{FG}}{{CG}}=\frac{{BG}}{{GD}}\:\: \\ $$$${GD}={BD}−{BG}=\mathrm{2}{R}\sqrt{\mathrm{2}}\:−\mathrm{3} \\ $$$${FB}={r}\:\:\:\:\:\:\:{CD}=\mathrm{2}{R} \\ $$$$\:\frac{{r}}{\mathrm{2}{R}}=\frac{\mathrm{3}}{\mathrm{2}{R}\sqrt{\mathrm{2}}\:−\mathrm{3}}\:\:\:\:\frac{\mathrm{1}}{\mathrm{3}}=\frac{\mathrm{3}}{\mathrm{2}{R}\sqrt{\mathrm{2}}\:−\mathrm{3}} \\ $$$$\Rightarrow\:\:\:{R}=\mathrm{3}\sqrt{\mathrm{2}}\:\:\:\:\:\:\:\:\:\left({i}\right) \\ $$$$\bigtriangleup{CDE}\:\:\mathrm{tan}\:\alpha=\frac{\mathrm{1}}{\mathrm{2}}\:\:\:{CE}={R}\sqrt{\mathrm{5}} \\ $$$$\bigtriangleup{DEH}\:\:\frac{\mathrm{cos}\:\alpha}{{DH}}=\frac{\mathrm{sin}\:\mathrm{45}}{{EH}}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{1}\right) \\ $$$$\bigtriangleup{CDH}\:\:\:\:\frac{\mathrm{sin}\:\alpha}{{DH}}=\frac{\mathrm{sin}\:\mathrm{45}}{{R}\sqrt{\mathrm{5}}\:−{EH}}\:\:\:\left(\mathrm{2}\right) \\ $$$$\mathrm{1}\:\:{EH}=\frac{{DH}\mathrm{sin}\:\mathrm{45}}{\mathrm{cos}\:\alpha} \\ $$$${DH}\mathrm{sin}\:\mathrm{45}=\left({R}\sqrt{\mathrm{5}}\:−{EH}\right)\mathrm{sin}\:\alpha \\ $$$${EH}=\frac{\mathrm{1}}{\mathrm{sin}\:\alpha}\left({R}\sqrt{\mathrm{5}\:}\:\mathrm{sin}\alpha−{DH}\mathrm{sin}\:\mathrm{45}\right) \\ $$$$\Rightarrow\frac{{DH}\mathrm{sin}\:\mathrm{45}}{\mathrm{cos}\:\alpha}=\frac{{R}\sqrt{\mathrm{5}}\:\mathrm{sin}\:\alpha−{DH}\mathrm{sin}\:\mathrm{45}}{\mathrm{sin}\:\alpha} \\ $$$$\mathrm{tan}\:\alpha=\frac{{R}\sqrt{\mathrm{5}}\:\mathrm{sin}\:\alpha−{DH}\mathrm{sin}\:\mathrm{45}}{{DH}\mathrm{sin}\:\mathrm{45}} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}=\frac{{R}\sqrt{\mathrm{5}}\:×\frac{{R}}{{R}\sqrt{\mathrm{5}}}−\frac{{DH}\sqrt{\mathrm{2}}}{\mathrm{2}}}{\frac{{DH}\sqrt{\mathrm{2}}}{\mathrm{2}}}\Rightarrow{DH}=\frac{\mathrm{4}{R}}{\mathrm{3}\sqrt{\mathrm{2}}} \\ $$$$\left({i}\right){R}=\mathrm{3}\sqrt{\mathrm{2}}\:\:\:\:\Rightarrow\:\:\:\boldsymbol{{DH}}=\mathrm{4}\:\: \\ $$$$ \\ $$
Commented by a.lgnaoui last updated on 28/Mar/23
Answered by mr W last updated on 29/Mar/23
(R+r)^2 −R^2 =(2R−r)^2 ⇒2R=3r ((3/( (√2)))/r)=((2R−(3/( (√2))))/(2R))=((3r−(3/( (√2))))/(3r)) ⇒r=(4/( (√2))) ((x/( (√2)))/R)=((2R−(x/( (√2))))/(2R)) ⇒x=((2(√2)R)/3)=(√2)r=4 ✓
$$\left({R}+{r}\right)^{\mathrm{2}} −{R}^{\mathrm{2}} =\left(\mathrm{2}{R}−{r}\right)^{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{2}{R}=\mathrm{3}{r} \\ $$$$\frac{\frac{\mathrm{3}}{\:\sqrt{\mathrm{2}}}}{{r}}=\frac{\mathrm{2}{R}−\frac{\mathrm{3}}{\:\sqrt{\mathrm{2}}}}{\mathrm{2}{R}}=\frac{\mathrm{3}{r}−\frac{\mathrm{3}}{\:\sqrt{\mathrm{2}}}}{\mathrm{3}{r}} \\ $$$$\Rightarrow{r}=\frac{\mathrm{4}}{\:\sqrt{\mathrm{2}}} \\ $$$$\frac{\frac{{x}}{\:\sqrt{\mathrm{2}}}}{{R}}=\frac{\mathrm{2}{R}−\frac{{x}}{\:\sqrt{\mathrm{2}}}}{\mathrm{2}{R}} \\ $$$$\Rightarrow{x}=\frac{\mathrm{2}\sqrt{\mathrm{2}}{R}}{\mathrm{3}}=\sqrt{\mathrm{2}}{r}=\mathrm{4}\:\checkmark \\ $$

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