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Question Number 77330 by TawaTawa last updated on 05/Jan/20

Commented by $@ty@m123 last updated on 05/Jan/20

Commented by $@ty@m123 last updated on 05/Jan/20

$$In△PDC\mathrm{\&}△QAP$$$$\mathrm{\angle }PDC=\mathrm{\angle }PAQ(={90}^{\mathrm{o}})$$$$\mathrm{\angle }PC\mathrm{D}=\mathrm{\angle }APQ(Let\mathrm{\angle }PC\mathrm{D}=\alpha \Rightarrow \mathrm{\angle }DPC=90-\alpha$$$$\Rightarrow \mathrm{\angle }APQ=180-(90-\alpha )-90=\alpha )$$$$\therefore △PDC\sim △QAP$$$$\Rightarrow \frac{x-y}{x}=\frac{3}{4}(wherePD=y)$$$$\Rightarrow 3x=4x-4y$$$$\Rightarrow 4y=x$$$$\Rightarrow y=\frac{x}{4}...\left(1\right)$$$$In△PDC,$$$$x^2+y^2=4^2$$$$\Rightarrow x^2+{\left(\frac{x}{4}\right)}^2=16$$$$\Rightarrow x=\frac{16}{\sqrt{17}}$$

Commented by TawaTawa last updated on 05/Jan/20

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

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