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
$${In}\:\bigtriangleup{PDC}\:\&\bigtriangleup{QAP} \\ $$$$\angle{PDC}=\angle{PAQ}\:\left(=\mathrm{90}^{\mathrm{o}} \right) \\ $$$$\angle{PC}\mathrm{D}=\angle{APQ}\:\left({Let}\:\angle{PC}\mathrm{D}=\alpha\Rightarrow\angle{DPC}=\mathrm{90}−\alpha\right. \\ $$$$\left.\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\Rightarrow\angle{APQ}=\mathrm{180}−\left(\mathrm{90}−\alpha\right)−\mathrm{90}=\alpha\right) \\ $$$$\therefore\:\bigtriangleup{PDC}\:\sim\bigtriangleup{QAP} \\ $$$$\Rightarrow\frac{{x}−{y}}{{x}}=\frac{\mathrm{3}}{\mathrm{4}}\:\left({where}\:{PD}={y}\right) \\ $$$$\Rightarrow\mathrm{3}{x}=\mathrm{4}{x}−\mathrm{4}{y} \\ $$$$\Rightarrow\mathrm{4}{y}={x} \\ $$$$\Rightarrow{y}=\frac{{x}}{\mathrm{4}}\:…\left(\mathrm{1}\right) \\ $$$${In}\:\bigtriangleup{PDC}, \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\mathrm{4}^{\mathrm{2}} \\ $$$$\Rightarrow{x}^{\mathrm{2}} +\left(\frac{{x}}{\mathrm{4}}\right)^{\mathrm{2}} =\mathrm{16} \\ $$$$\Rightarrow{x}=\frac{\mathrm{16}}{\:\sqrt{\mathrm{17}}} \\ $$
Commented by TawaTawa last updated on 05/Jan/20
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$