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Question Number 199544 by a.lgnaoui last updated on 05/Nov/23

$$\frac{\mathbf{Area}\left(\mathit{y}\mathit{d}\mathit{l}\mathit{l}\mathit{o}\mathit{w}\right)}{\mathbf{Area}\left(\mathit{B}\mathit{l}\mathit{u}\mathit{e}\right)}=?$$

Commented by a.lgnaoui last updated on 05/Nov/23

Commented by ajfour last updated on 05/Nov/23

$$x\cos \theta +y\sin \theta =1$$$$fory=1,x=\frac{1}{2}$$$$\frac{\cos \theta }{2}+\sin \theta =1$$$$\Rightarrow 4-4\cos {}^2\theta ={(2-\cos \theta )}^2$$$$\Rightarrow 5\cos {}^2\theta -4\cos \theta =0$$$$\cos \theta =\frac{4}{5}\Rightarrow \tan \theta =\frac{3}{4}$$$$A_y=\frac{1}{2}{\left(\frac{1}{2}\right)}^2\left(\frac{\cos \theta }{\sin \theta }\right)$$$$=\frac{1}{8}\times \frac{4}{3}=\frac{1}{6}$$$$\frac{A_y}{A_b}=\frac{A_y}{1-A_y}=\frac{1/6}{5/6}=\frac{1}{5}$$$$$$

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

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

$$saysidelengthofsquareis1.$$$${(1-x)}^2+{\left(\frac{1}{2}\right)}^2={(x+\frac{1}{2})}^2$$$$\Rightarrow x=\frac{1}{3}$$$$areayellow=\frac{1}{2}\times \frac{1}{2}\times (1-\frac{1}{3})=\frac{1}{6}$$$$areablue=1^2-yellow=1-\frac{1}{6}=\frac{5}{6}$$$$\Rightarrow \frac{yellow}{blue}=\frac{1}{5}✓$$

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