Question Number 211016 by mr W last updated on 26/Aug/24
Commented by mr W last updated on 26/Aug/24
$${find}\:{x}=? \\ $$$$ \\ $$$${the}\:{original}\:{question}\:{was}\:{deleted}\:{by} \\ $$$${the}\:{questioner}\:{for}\:{some}\:{reasons}. \\ $$
Answered by mr W last updated on 26/Aug/24
Commented by mr W last updated on 26/Aug/24
$${Method}\:{II} \\ $$$${FB}={AD}={x}={FE} \\ $$$$\Rightarrow{BG}={GE}=\frac{\mathrm{5}+\mathrm{5}+\mathrm{2}}{\mathrm{2}}=\mathrm{6} \\ $$$${GC}=\mathrm{6}−\mathrm{2}=\mathrm{4} \\ $$$${FG}=\sqrt{\mathrm{5}^{\mathrm{2}} −\mathrm{4}^{\mathrm{2}} }=\mathrm{3} \\ $$$${x}=\sqrt{\mathrm{6}^{\mathrm{2}} +\mathrm{3}^{\mathrm{2}} }=\mathrm{3}\sqrt{\mathrm{5}} \\ $$
Answered by A5T last updated on 26/Aug/24
$${Let}\:{AE}={y} \\ $$$$\Rightarrow\mathrm{10}×\mathrm{5}^{\mathrm{2}} +\mathrm{10}{x}^{\mathrm{2}} =\mathrm{2}^{\mathrm{2}} ×\mathrm{5}+{y}^{\mathrm{2}} ×\mathrm{5} \\ $$$$\Rightarrow\mathrm{50}+\mathrm{2}{x}^{\mathrm{2}} =\mathrm{4}+{y}^{\mathrm{2}} \Rightarrow{y}^{\mathrm{2}} −\mathrm{2}{x}^{\mathrm{2}} =\mathrm{46}…\left({i}\right) \\ $$$$\mathrm{5}×\mathrm{2}×\mathrm{7}+\mathrm{10}^{\mathrm{2}} ×\mathrm{7}=\mathrm{5}{y}^{\mathrm{2}} +\mathrm{2}{x}^{\mathrm{2}} \Rightarrow\mathrm{5}{y}^{\mathrm{2}} +\mathrm{2}{x}^{\mathrm{2}} =\mathrm{770}…\left({ii}\right) \\ $$$$\left({ii}\right)+\left({i}\right)\Rightarrow\mathrm{6}{y}^{\mathrm{2}} =\mathrm{816}\Rightarrow{y}^{\mathrm{2}} =\mathrm{136}\Rightarrow{x}=\mathrm{3}\sqrt{\mathrm{5}} \\ $$
Commented by mr W last updated on 26/Aug/24
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Answered by mr W last updated on 26/Aug/24
Commented by mr W last updated on 26/Aug/24
$${METHOD}\:{I} \\ $$$$\mathrm{2}{y}^{\mathrm{2}} +\mathrm{5}{x}^{\mathrm{2}} =\mathrm{7}\left(\mathrm{5}^{\mathrm{2}} +\mathrm{5}×\mathrm{2}\right)\:\:\:…\left({i}\right) \\ $$$$\mathrm{10}^{\mathrm{2}} +\mathrm{4}{y}^{\mathrm{2}} =\mathrm{2}\left({x}^{\mathrm{2}} +\mathrm{5}×\mathrm{5}\right)\:\:\:…\left({ii}\right) \\ $$$$\left({ii}\right)−\mathrm{2}×\left({i}\right): \\ $$$$\mathrm{10}^{\mathrm{2}} −\mathrm{10}{x}^{\mathrm{2}} =\mathrm{2}\left({x}^{\mathrm{2}} +\mathrm{25}\right)−\mathrm{14}×\mathrm{35} \\ $$$$\Rightarrow{x}=\mathrm{3}\sqrt{\mathrm{5}} \\ $$