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Question-166834

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Question Number 166834 by leicianocosta last updated on 28/Feb/22
Answered by Rasheed.Sindhi last updated on 28/Feb/22
Commented by Rasheed.Sindhi last updated on 28/Feb/22
Let AC=BC=a AC^2 +BC^2 =AB^2 a^2 +a^2 =6^2 ⇒a=3(√2) CD^2 =AC^2 +AD^2 −2.AC.AD.cos A =(3(√2) )^2 +2^2 −2.3(√2) .2.cos 45 =18+4−12(√2) .(1/( (√2) ))=10 CD=(√(10)) △BCD: ((BD)/(sin C))=((CD)/(sin B))⇒(4/(sin C))=(((√(10)) )/(sin 45)) ⇒(√(10)) sin C=4((1/( (√2)))) sin C=(4/( (√(10)) .(√2)))=(2/( (√5))) C=sin^(−1) ((2/( (√5) ))) D=180−(45+sin^(−1) ((2/( (√5))))) BDE=D−45 =180−45−sin^(−1) ((2/( (√5))))−45 =90−sin^(−1) ((2/( (√5)))) △BDE: ∠BDE=90−sin^(−1) ((2/( (√5)))) BD=4 ∠B=45 ∠E=180−(45+90−sin^(−1) ((2/( (√5))))) =45+sin^(−1) ((2/( (√5)))) ▲BDE=((BD^2 .sinB.sinD )/(2 sinE )) =((4^2 sin45.sin(90−sin^(−1) ((2/( (√5))))) )/(2(45+sin^(−1) ((2/( (√5))))))) =((((16)/( (√2))).cos(sin^(−1) ((2/( (√5))))) )/(90+2sin^(−1) ((2/( (√5)))))))
$${Let}\:{AC}={BC}={a} \\ $$$${AC}^{\mathrm{2}} +{BC}^{\mathrm{2}} ={AB}^{\mathrm{2}} \\ $$$$\:{a}^{\mathrm{2}} +{a}^{\mathrm{2}} =\mathrm{6}^{\mathrm{2}} \Rightarrow{a}=\mathrm{3}\sqrt{\mathrm{2}}\: \\ $$$${CD}^{\mathrm{2}} ={AC}^{\mathrm{2}} +{AD}^{\mathrm{2}} −\mathrm{2}.{AC}.{AD}.\mathrm{cos}\:{A} \\ $$$$\:\:\:\:\:\:\:\:\:=\left(\mathrm{3}\sqrt{\mathrm{2}}\:\right)^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} −\mathrm{2}.\mathrm{3}\sqrt{\mathrm{2}}\:.\mathrm{2}.\mathrm{cos}\:\mathrm{45} \\ $$$$\:\:\:\:\:\:\:\:\:=\mathrm{18}+\mathrm{4}−\mathrm{12}\sqrt{\mathrm{2}}\:.\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}\:}=\mathrm{10} \\ $$$${CD}=\sqrt{\mathrm{10}}\: \\ $$$$\bigtriangleup{BCD}: \\ $$$$\frac{{BD}}{\mathrm{sin}\:{C}}=\frac{{CD}}{\mathrm{sin}\:{B}}\Rightarrow\frac{\mathrm{4}}{\mathrm{sin}\:{C}}=\frac{\sqrt{\mathrm{10}}\:}{\mathrm{sin}\:\mathrm{45}} \\ $$$$\Rightarrow\sqrt{\mathrm{10}}\:\mathrm{sin}\:{C}=\mathrm{4}\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\right) \\ $$$$\:\:\:\:\:\:\:\mathrm{sin}\:{C}=\frac{\mathrm{4}}{\:\sqrt{\mathrm{10}}\:.\sqrt{\mathrm{2}}}=\frac{\mathrm{2}}{\:\sqrt{\mathrm{5}}} \\ $$$$\:\:\:\:\:\:\:\:\:{C}=\mathrm{sin}^{−\mathrm{1}} \left(\frac{\mathrm{2}}{\:\sqrt{\mathrm{5}}\:}\right) \\ $$$$\:\:\:\:\:\:\:\:\:{D}=\mathrm{180}−\left(\mathrm{45}+\mathrm{sin}^{−\mathrm{1}} \left(\frac{\mathrm{2}}{\:\sqrt{\mathrm{5}}}\right)\right) \\ $$$$\:\:\:\:\:{BDE}={D}−\mathrm{45} \\ $$$$\:\:\:\:=\mathrm{180}−\mathrm{45}−\mathrm{sin}^{−\mathrm{1}} \left(\frac{\mathrm{2}}{\:\sqrt{\mathrm{5}}}\right)−\mathrm{45} \\ $$$$\:\:\:\:=\mathrm{90}−\mathrm{sin}^{−\mathrm{1}} \left(\frac{\mathrm{2}}{\:\sqrt{\mathrm{5}}}\right) \\ $$$$\bigtriangleup{BDE}: \\ $$$$\:\:\:\:\angle{BDE}=\mathrm{90}−\mathrm{sin}^{−\mathrm{1}} \left(\frac{\mathrm{2}}{\:\sqrt{\mathrm{5}}}\right) \\ $$$$\:\:\:{BD}=\mathrm{4} \\ $$$$\:\:\:\angle{B}=\mathrm{45} \\ $$$$\:\:\:\:\angle{E}=\mathrm{180}−\left(\mathrm{45}+\mathrm{90}−\mathrm{sin}^{−\mathrm{1}} \left(\frac{\mathrm{2}}{\:\sqrt{\mathrm{5}}}\right)\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{45}+\mathrm{sin}^{−\mathrm{1}} \left(\frac{\mathrm{2}}{\:\sqrt{\mathrm{5}}}\right) \\ $$$$\blacktriangle{BDE}=\frac{{BD}^{\mathrm{2}} .\mathrm{sin}{B}.\mathrm{sin}{D}\:\:}{\mathrm{2}\:\mathrm{sin}{E}\:} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{4}^{\mathrm{2}} \mathrm{sin45}.\mathrm{sin}\left(\mathrm{90}−\mathrm{sin}^{−\mathrm{1}} \left(\frac{\mathrm{2}}{\:\sqrt{\mathrm{5}}}\right)\right)\:\:}{\mathrm{2}\left(\mathrm{45}+\mathrm{sin}^{−\mathrm{1}} \left(\frac{\mathrm{2}}{\:\sqrt{\mathrm{5}}}\right)\right)} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\frac{\mathrm{16}}{\:\sqrt{\mathrm{2}}}.\mathrm{cos}\left(\mathrm{sin}^{−\mathrm{1}} \left(\frac{\mathrm{2}}{\:\sqrt{\mathrm{5}}}\right)\right)\:\:}{\left.\mathrm{90}+\mathrm{2sin}^{−\mathrm{1}} \left(\frac{\mathrm{2}}{\:\sqrt{\mathrm{5}}}\right)\right)} \\ $$
Commented by Tawa11 last updated on 01/Mar/22
Great sir
$$\mathrm{Great}\:\mathrm{sir} \\ $$
Answered by mr W last updated on 28/Feb/22
Commented by mr W last updated on 28/Feb/22
AC=3(√2) CD^2 =((2/( (√2))))^2 +(3(√2)−(2/( (√2))))^2 =10 ⇒CD=(√(10)) ((sin α)/2)=((sin 45°)/( (√(10)))) sin α=(1/( (√5))) ⇒cos α=(2/( (√5))) DF=FE CE×cos α+CE×sin α=CD CE=((√(10))/((1/( (√5)))+(2/( (√5)))))=((5(√2))/3) BE=3(√2)−((5(√2))/3)=((4(√2))/3) x=(1/2)×4×((4(√2))/3)×((√2)/2)=(8/3)
$${AC}=\mathrm{3}\sqrt{\mathrm{2}} \\ $$$${CD}^{\mathrm{2}} =\left(\frac{\mathrm{2}}{\:\sqrt{\mathrm{2}}}\right)^{\mathrm{2}} +\left(\mathrm{3}\sqrt{\mathrm{2}}−\frac{\mathrm{2}}{\:\sqrt{\mathrm{2}}}\right)^{\mathrm{2}} =\mathrm{10}\:\Rightarrow{CD}=\sqrt{\mathrm{10}} \\ $$$$\frac{\mathrm{sin}\:\alpha}{\mathrm{2}}=\frac{\mathrm{sin}\:\mathrm{45}°}{\:\sqrt{\mathrm{10}}} \\ $$$$\mathrm{sin}\:\alpha=\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}\:\Rightarrow\mathrm{cos}\:\alpha=\frac{\mathrm{2}}{\:\sqrt{\mathrm{5}}} \\ $$$${DF}={FE} \\ $$$${CE}×\mathrm{cos}\:\alpha+{CE}×\mathrm{sin}\:\alpha={CD} \\ $$$${CE}=\frac{\sqrt{\mathrm{10}}}{\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}+\frac{\mathrm{2}}{\:\sqrt{\mathrm{5}}}}=\frac{\mathrm{5}\sqrt{\mathrm{2}}}{\mathrm{3}} \\ $$$${BE}=\mathrm{3}\sqrt{\mathrm{2}}−\frac{\mathrm{5}\sqrt{\mathrm{2}}}{\mathrm{3}}=\frac{\mathrm{4}\sqrt{\mathrm{2}}}{\mathrm{3}} \\ $$$${x}=\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{4}×\frac{\mathrm{4}\sqrt{\mathrm{2}}}{\mathrm{3}}×\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}=\frac{\mathrm{8}}{\mathrm{3}} \\ $$
Commented by Tawa11 last updated on 01/Mar/22
Great sir
$$\mathrm{Great}\:\mathrm{sir} \\ $$

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