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Question Number 199544 by a.lgnaoui last updated on 05/Nov/23
((Area(ydllow))/(Area(Blue)))=?
$$\:\:\frac{\boldsymbol{\mathrm{Area}}\left(\boldsymbol{{ydllow}}\right)}{\boldsymbol{\mathrm{Area}}\left(\boldsymbol{{Blue}}\right)}=? \\ $$
Commented by a.lgnaoui last updated on 05/Nov/23
Commented by ajfour last updated on 05/Nov/23
xcos θ+ysin θ=1 for y=1 , x=(1/2) ((cos θ)/2)+sin θ=1 ⇒ 4−4cos^2 θ=(2−cos θ)^2 ⇒ 5cos^2 θ−4cos θ=0 cos θ=(4/5) ⇒ tan θ=(3/4) A_y =(1/2)((1/2))^2 (((cos θ)/(sin θ))) = (1/8)×(4/3)=(1/6) (A_y /A_b )=(A_y /(1−A_y ))=((1/6)/(5/6))=(1/5)
$${x}\mathrm{cos}\:\theta+{y}\mathrm{sin}\:\theta=\mathrm{1} \\ $$$${for}\:{y}=\mathrm{1}\:,\:\:{x}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\frac{\mathrm{cos}\:\theta}{\mathrm{2}}+\mathrm{sin}\:\theta=\mathrm{1} \\ $$$$\Rightarrow\:\:\mathrm{4}−\mathrm{4cos}\:^{\mathrm{2}} \theta=\left(\mathrm{2}−\mathrm{cos}\:\theta\right)^{\mathrm{2}} \\ $$$$\Rightarrow\:\:\mathrm{5cos}\:^{\mathrm{2}} \theta−\mathrm{4cos}\:\theta=\mathrm{0} \\ $$$$\mathrm{cos}\:\theta=\frac{\mathrm{4}}{\mathrm{5}}\:\:\:\:\Rightarrow\:\mathrm{tan}\:\theta=\frac{\mathrm{3}}{\mathrm{4}} \\ $$$${A}_{{y}} =\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} \left(\frac{\mathrm{cos}\:\theta}{\mathrm{sin}\:\theta}\right) \\ $$$$\:\:\:\:\:\:=\:\frac{\mathrm{1}}{\mathrm{8}}×\frac{\mathrm{4}}{\mathrm{3}}=\frac{\mathrm{1}}{\mathrm{6}} \\ $$$$\frac{{A}_{{y}} }{{A}_{{b}} }=\frac{{A}_{{y}} }{\mathrm{1}−{A}_{{y}} }=\frac{\mathrm{1}/\mathrm{6}}{\mathrm{5}/\mathrm{6}}=\frac{\mathrm{1}}{\mathrm{5}} \\ $$$$ \\ $$
Answered by mr W last updated on 05/Nov/23
Commented by mr W last updated on 05/Nov/23
say side length of square is 1. (1−x)^2 +((1/2))^2 =(x+(1/2))^2 ⇒x=(1/3) area yellow =(1/2)×(1/2)×(1−(1/3))=(1/6) area blue =1^2 −yellow=1−(1/6)=(5/6) ⇒((yellow)/(blue))=(1/5) ✓
$${say}\:{side}\:{length}\:{of}\:{square}\:{is}\:\mathrm{1}. \\ $$$$\left(\mathrm{1}−{x}\right)^{\mathrm{2}} +\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} =\left({x}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} \\ $$$$\Rightarrow{x}=\frac{\mathrm{1}}{\mathrm{3}} \\ $$$${area}\:{yellow}\:=\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{1}}{\mathrm{2}}×\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{3}}\right)=\frac{\mathrm{1}}{\mathrm{6}} \\ $$$${area}\:{blue}\:=\mathrm{1}^{\mathrm{2}} −{yellow}=\mathrm{1}−\frac{\mathrm{1}}{\mathrm{6}}=\frac{\mathrm{5}}{\mathrm{6}} \\ $$$$\Rightarrow\frac{{yellow}}{{blue}}=\frac{\mathrm{1}}{\mathrm{5}}\:\checkmark \\ $$

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