Question Number 223124 by Jubr last updated on 15/Jul/25
Answered by fantastic last updated on 15/Jul/25
Commented by fantastic last updated on 15/Jul/25
$${EB}={x}\sqrt{\mathrm{2}}={EC} \\ $$$${AB}=\sqrt{\left({x}\sqrt{\mathrm{2}}\right)^{\mathrm{2}} −\mathrm{8}^{\mathrm{2}} }=\sqrt{\mathrm{2}{x}^{\mathrm{2}} −\mathrm{8}^{\mathrm{2}} } \\ $$$${CD}=\sqrt{\left({x}\sqrt{\mathrm{2}}\right)^{\mathrm{2}} −\mathrm{6}^{\mathrm{2}} }=\sqrt{\mathrm{2}{x}^{\mathrm{2}} −\mathrm{6}^{\mathrm{2}} } \\ $$$$\bigtriangleup{ABE}\sim\bigtriangleup{CDE} \\ $$$$\frac{\mathrm{8}}{\:\sqrt{\mathrm{2}{x}^{\mathrm{2}} −\mathrm{6}^{\mathrm{2}} }}=\frac{\mathrm{6}}{\:\sqrt{\mathrm{2}{x}^{\mathrm{2}} −\mathrm{8}^{\mathrm{2}} }} \\ $$$${or}\:\frac{\mathrm{4}}{\mathrm{3}}=\frac{\sqrt{\mathrm{2}{x}^{\mathrm{2}} −\mathrm{6}^{\mathrm{2}} }}{\:\sqrt{\mathrm{2}{x}^{\mathrm{2}} −\mathrm{8}^{\mathrm{2}} }} \\ $$$${or}\:\frac{\mathrm{16}}{\mathrm{9}}=\frac{\mathrm{2}{x}^{\mathrm{2}} −\mathrm{6}^{\mathrm{2}} }{\mathrm{2}{x}^{\mathrm{2}} −\mathrm{8}^{\mathrm{2}} } \\ $$$${or}\:\frac{\mathrm{16}}{\mathrm{9}}=\frac{{x}^{\mathrm{2}} −\mathrm{18}}{{x}^{\mathrm{2}} −\mathrm{32}} \\ $$$${or}\:\mathrm{16}{x}^{\mathrm{2}} −\mathrm{16}×\mathrm{32}=\mathrm{9}{x}^{\mathrm{2}} −\mathrm{9}×\mathrm{18} \\ $$$${or}\:\mathrm{7}{x}^{\mathrm{2}} =\mathrm{350} \\ $$$${x}^{\mathrm{2}} =\mathrm{50} \\ $$$${x}=\sqrt{\mathrm{50}}=\mathrm{5}\sqrt{\mathrm{2}} \\ $$
Commented by Tawa11 last updated on 15/Jul/25
$$\mathrm{Sir},\:\mathrm{please}\:\mathrm{how}\:\mathrm{is}\:\mathrm{the}\:\mathrm{triangle}\:\mathrm{similar}? \\ $$$$\mathrm{I}\:\mathrm{think}\:\mathrm{they}\:\mathrm{should}\:\mathrm{have}\:\mathrm{common}\:\mathrm{vertex}. \\ $$$$\mathrm{Just}\:\mathrm{asking}\:\mathrm{sir}. \\ $$
Commented by fantastic last updated on 15/Jul/25
$${let}\:\measuredangle{ABE}=\theta\:{then}\:\measuredangle{AEB}=\mathrm{90}^{\mathrm{0}} −\theta \\ $$$${Now}\:\bigtriangleup{BFE}\:{and}\bigtriangleup{ECF}\:{are}\:\:{iso}\:{celes}\:{triangles} \\ $$$$\measuredangle{BEF}=\mathrm{45}^{\mathrm{0}} =\measuredangle{CEF} \\ $$$${So}\:\measuredangle{BEC}=\mathrm{45}^{\mathrm{0}} +\mathrm{45}^{\mathrm{0}} =\mathrm{90}^{\mathrm{0}} \: \\ $$$${So}\:\measuredangle{DEC}=\mathrm{180}^{\mathrm{0}} −\left(\mathrm{90}^{\mathrm{0}} −\theta+\mathrm{90}^{\mathrm{0}} \right)=\theta \\ $$$${So}\:\measuredangle{ABE}=\measuredangle{DEC} \\ $$$${We}\:{have}\:\measuredangle{A}=\measuredangle{D}=\mathrm{90}^{\mathrm{0}} \\ $$$${So}\:\measuredangle{ECD}=\mathrm{90}^{\mathrm{0}} −\theta=\measuredangle{AEB} \\ $$$${So}\:\bigtriangleup{AEB}\sim\bigtriangleup{CDE} \\ $$
Commented by Tawa11 last updated on 16/Jul/25
$$\mathrm{I}\:\mathrm{understand}\:\mathrm{now}\:\mathrm{sir}.\:\mathrm{Thanks}\:\mathrm{sir} \\ $$
Answered by mr W last updated on 15/Jul/25
Commented by mr W last updated on 15/Jul/25
$${x}=\sqrt{\mathrm{1}^{\mathrm{2}} +\mathrm{7}^{\mathrm{2}} }=\mathrm{5}\sqrt{\mathrm{2}}\:\checkmark \\ $$
Commented by Tawa11 last updated on 16/Jul/25
$$\mathrm{Very}\:\mathrm{good}\:\mathrm{method}. \\ $$
Answered by mr W last updated on 15/Jul/25
Commented by mr W last updated on 15/Jul/25
$$\Delta{ABE}\equiv\Delta{CED} \\ $$$${AB}={CE}=\mathrm{6} \\ $$$$\left(\sqrt{\mathrm{2}}{x}\right)^{\mathrm{2}} =\mathrm{8}^{\mathrm{2}} +\mathrm{6}^{\mathrm{2}} \:\Rightarrow{x}=\mathrm{5}\sqrt{\mathrm{2}}\:\checkmark \\ $$
Commented by Tawa11 last updated on 16/Jul/25
$$\mathrm{Weldone}\:\mathrm{sir}. \\ $$