Question-218493

Question Number 218493 by Spillover last updated on 10/Apr/25

Answered by mr W last updated on 11/Apr/25

Commented by mr W last updated on 11/Apr/25

s=side length of largest square tan α=(1/2) ⇒sin α=(1/( (√5))), cos α=(2/( (√5))) s=(x/(cos α))+x sin α s=x cos α+(3/(cos α))+3 sin α x cos α+(3/(cos α))+3 sin α=(x/(cos α))+x sin α ⇒x=((3((1/(cos α))+sin α))/((1/(cos α))+sin α−cos α)) =((3(((√5)/2)+(1/( (√5)))))/(((√5)/2)+(1/( (√5)))−(2/( (√5)))))=7 area ?=x^2 =7^2 =49

$${s}={side}\:{length}\:{of}\:{largest}\:{square} \\ $$$$\mathrm{tan}\:\alpha=\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow\mathrm{sin}\:\alpha=\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}},\:\mathrm{cos}\:\alpha=\frac{\mathrm{2}}{\:\sqrt{\mathrm{5}}} \\ $$$${s}=\frac{{x}}{\mathrm{cos}\:\alpha}+{x}\:\mathrm{sin}\:\alpha \\ $$$${s}={x}\:\mathrm{cos}\:\alpha+\frac{\mathrm{3}}{\mathrm{cos}\:\alpha}+\mathrm{3}\:\mathrm{sin}\:\alpha \\ $$$${x}\:\mathrm{cos}\:\alpha+\frac{\mathrm{3}}{\mathrm{cos}\:\alpha}+\mathrm{3}\:\mathrm{sin}\:\alpha=\frac{{x}}{\mathrm{cos}\:\alpha}+{x}\:\mathrm{sin}\:\alpha \\ $$$$\Rightarrow{x}=\frac{\mathrm{3}\left(\frac{\mathrm{1}}{\mathrm{cos}\:\alpha}+\mathrm{sin}\:\alpha\right)}{\frac{\mathrm{1}}{\mathrm{cos}\:\alpha}+\mathrm{sin}\:\alpha−\mathrm{cos}\:\alpha} \\ $$$$\:\:\:\:\:\:=\frac{\mathrm{3}\left(\frac{\sqrt{\mathrm{5}}}{\mathrm{2}}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}\right)}{\frac{\sqrt{\mathrm{5}}}{\mathrm{2}}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}−\frac{\mathrm{2}}{\:\sqrt{\mathrm{5}}}}=\mathrm{7} \\ $$$${area}\:?={x}^{\mathrm{2}} =\mathrm{7}^{\mathrm{2}} =\mathrm{49} \\ $$

Commented by Spillover last updated on 11/Apr/25

$${correct} \\ $$

Answered by Spillover last updated on 11/Apr/25

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