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Question Number 184352 by cherokeesay last updated on 05/Jan/23

Answered by som(math1967) last updated on 05/Jan/23

△ABC∼△ACD ⇒x^2 =6×4⇒x=(√(24))=2(√6) BC=(√(4×2))=2(√2) △AFE∼△ABC ((x+y)/4)=(x/(2(√2))) 2(√6)+y=4×((2(√6))/(2(√2))) y=((4(√6)−2(√(12)))/( (√2)))=((4(√6)−4(√3))/( (√2))) (x/y)=((2(√6)×(√2))/(4((√6)−(√3))))=((√3)/( (√3)((√2)−1)))=(√2)+1

$$\bigtriangleup{ABC}\sim\bigtriangleup{ACD} \\ $$$$\:\Rightarrow{x}^{\mathrm{2}} =\mathrm{6}×\mathrm{4}\Rightarrow{x}=\sqrt{\mathrm{24}}=\mathrm{2}\sqrt{\mathrm{6}} \\ $$$${BC}=\sqrt{\mathrm{4}×\mathrm{2}}=\mathrm{2}\sqrt{\mathrm{2}} \\ $$$$\bigtriangleup{AFE}\sim\bigtriangleup{ABC} \\ $$$$\:\frac{{x}+{y}}{\mathrm{4}}=\frac{{x}}{\mathrm{2}\sqrt{\mathrm{2}}} \\ $$$$\mathrm{2}\sqrt{\mathrm{6}}+{y}=\mathrm{4}×\frac{\mathrm{2}\sqrt{\mathrm{6}}}{\mathrm{2}\sqrt{\mathrm{2}}} \\ $$$$\:\:{y}=\frac{\mathrm{4}\sqrt{\mathrm{6}}−\mathrm{2}\sqrt{\mathrm{12}}}{\:\sqrt{\mathrm{2}}}=\frac{\mathrm{4}\sqrt{\mathrm{6}}−\mathrm{4}\sqrt{\mathrm{3}}}{\:\sqrt{\mathrm{2}}} \\ $$$$\:\:\frac{{x}}{{y}}=\frac{\mathrm{2}\sqrt{\mathrm{6}}×\sqrt{\mathrm{2}}}{\mathrm{4}\left(\sqrt{\mathrm{6}}−\sqrt{\mathrm{3}}\right)}=\frac{\sqrt{\mathrm{3}}}{\:\sqrt{\mathrm{3}}\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)}=\sqrt{\mathrm{2}}+\mathrm{1} \\ $$

Commented by som(math1967) last updated on 05/Jan/23

Commented by cherokeesay last updated on 05/Jan/23

nice ! thank you sir.

$${nice}\:!\:{thank}\:{you}\:{sir}. \\ $$

Answered by mr W last updated on 05/Jan/23

Commented by mr W last updated on 05/Jan/23

h^2 =4×2 ⇒h=2(√2) k=4−h=4−2(√2) (x/y)=(h/k)=((2(√2))/(4−2(√2)))=(1/( (√2)−1))=(√2)+1

$${h}^{\mathrm{2}} =\mathrm{4}×\mathrm{2}\:\Rightarrow{h}=\mathrm{2}\sqrt{\mathrm{2}} \\ $$$${k}=\mathrm{4}−{h}=\mathrm{4}−\mathrm{2}\sqrt{\mathrm{2}} \\ $$$$\frac{{x}}{{y}}=\frac{{h}}{{k}}=\frac{\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{4}−\mathrm{2}\sqrt{\mathrm{2}}}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}−\mathrm{1}}=\sqrt{\mathrm{2}}+\mathrm{1} \\ $$

Commented by cherokeesay last updated on 05/Jan/23

thank you sir.

$${thank}\:{you}\:{sir}. \\ $$

Answered by HeferH last updated on 05/Jan/23

h^2 = 4∙2 ⇒ h = 2(√2) (4/(4−2(√2))) = ((x + y)/y) (similar triangles) (2/(2 − (√2))) − 1= (x/y) ⇒ (x/y) = ((√2)/(2−(√2))) =(((2+(√2))(√2))/2) = ((2(√2) + 2)/2) =1 + (√2)

$${h}^{\mathrm{2}} =\:\mathrm{4}\centerdot\mathrm{2}\:\Rightarrow\:{h}\:=\:\mathrm{2}\sqrt{\mathrm{2}} \\ $$$$\:\frac{\mathrm{4}}{\mathrm{4}−\mathrm{2}\sqrt{\mathrm{2}}}\:=\:\frac{{x}\:+\:{y}}{{y}}\:\left({similar}\:{triangles}\right) \\ $$$$\:\frac{\mathrm{2}}{\mathrm{2}\:−\:\sqrt{\mathrm{2}}}\:\:−\:\mathrm{1}=\:\frac{{x}}{{y}}\:\Rightarrow\: \\ $$$$\:\frac{{x}}{{y}}\:=\:\frac{\sqrt{\mathrm{2}}}{\mathrm{2}−\sqrt{\mathrm{2}}}\:=\frac{\left(\mathrm{2}+\sqrt{\mathrm{2}}\right)\sqrt{\mathrm{2}}}{\mathrm{2}}\:=\:\frac{\mathrm{2}\sqrt{\mathrm{2}}\:+\:\mathrm{2}}{\mathrm{2}}\:=\mathrm{1}\:+\:\sqrt{\mathrm{2}}\: \\ $$

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