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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 &△QAP ∠PDC=∠PAQ (=90^o ) ∠PCD=∠APQ (Let ∠PCD=α⇒∠DPC=90−α ⇒∠APQ=180−(90−α)−90=α) ∴ △PDC ∼△QAP ⇒((x−y)/x)=(3/4) (where PD=y) ⇒3x=4x−4y ⇒4y=x ⇒y=(x/4) ...(1) In △PDC, x^2 +y^2 =4^2 ⇒x^2 +((x/4))^2 =16 ⇒x=((16)/(√(17)))

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

Commented by TawaTawa last updated on 05/Jan/20

God bless you sir

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

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