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Question Number 61559 by Tawa1 last updated on 04/Jun/19

Commented by Tawa1 last updated on 04/Jun/19

$$\mathrm{Find}\mathrm{the}\mathrm{length}\mathrm{of}\mathrm{the}\mathrm{pink}$$

Answered by ajfour last updated on 04/Jun/19

$$x^2+y^2=2\left(rightcircle\right)$$$${(x+\sqrt{2}+1)}^2+{(y-1)}^2=2\left(leftone\right)$$$$\Rightarrow 2(\sqrt{2}+1)x+3+2\sqrt{2}-2y+1=0$$$$\Rightarrow y=(\sqrt{2}+1)x+\sqrt{2}+2$$$$\Rightarrow x^2+{[(\sqrt{2}+1)x+\sqrt{2}+2]}^2=2$$$$\Rightarrow (4+2\sqrt{2})x^2+2\sqrt{2}(3+2\sqrt{2})x+4\sqrt{2}=0$$$$onerootisx=-\sqrt{2}$$$$\Rightarrow otherroot=-\frac{2\sqrt{2}{(\sqrt{2}+1)}^2}{2\sqrt{2}(\sqrt{2}+1)}+\sqrt{2}$$$$=-\sqrt{2}-1+\sqrt{2}=-1$$$$y=1$$$$pinksegmentlength=2$$

Commented by Tawa1 last updated on 04/Jun/19

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}$$

Answered by mr W last updated on 04/Jun/19

Commented by mr W last updated on 04/Jun/19

$$\mathrm{\angle }DCA=\mathrm{\angle }DCB$$$$\alpha =\frac{135°}{2}=67.5°$$$$\beta =180°-2\alpha =45°$$$$\gamma =135°-\beta =90°$$$$DE=\sqrt{2}AE=\sqrt{2}\times \sqrt{2}=2$$$$$$$$or$$$$BCADislozenge,\Rightarrow AD//BC$$$$\Rightarrow \beta =180-135=45°$$$$......$$

Commented by Tawa1 last updated on 04/Jun/19

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}$$

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