Question Number 226705 by mr W last updated on 10/Dec/25
Answered by fantastic2 last updated on 11/Dec/25
Commented by fantastic2 last updated on 11/Dec/25
$$\bigtriangleup{OPQ}\cong\bigtriangleup{SRQ} \\ $$$$\frac{{x}−\mathrm{5}}{\mathrm{2}}=\frac{\mathrm{5}}{{x}−\mathrm{2}} \\ $$$$\Rightarrow{x}^{\mathrm{2}} −\mathrm{7}{x}+\mathrm{10}=\mathrm{10} \\ $$$$\Rightarrow{x}=\mathrm{0}\:{or}\:{x}=\mathrm{7} \\ $$$$\therefore{x}=\mathrm{7}\checkmark \\ $$
Commented by fantastic2 last updated on 11/Dec/25
$${or} \\ $$$$\angle{OQP}+\angle{SQR}=\mathrm{90}^{\mathrm{0}} \\ $$$$\therefore\angle{OQS}=\mathrm{90}^{\mathrm{0}} \\ $$$${OS}=\sqrt{\mathrm{3}^{\mathrm{2}} +\mathrm{7}^{\mathrm{2}} }=\sqrt{\mathrm{58}} \\ $$$${OQ}=\sqrt{\mathrm{5}^{\mathrm{2}} +\left({x}−\mathrm{5}\right)^{\mathrm{2}} } \\ $$$${SQ}=\sqrt{\mathrm{2}^{\mathrm{2}} +\left({x}−\mathrm{2}\right)^{\mathrm{2}} } \\ $$$${OS}^{\mathrm{2}} ={OQ}^{\mathrm{2}} +{SQ}^{\mathrm{2}} \\ $$$${x}^{\mathrm{2}} −\mathrm{10}{x}+\mathrm{25}+{x}^{\mathrm{2}} −\mathrm{4}{x}+\mathrm{4}=\mathrm{58}−\mathrm{25}−\mathrm{4}=\mathrm{29} \\ $$$$\mathrm{2}{x}^{\mathrm{2}} −\mathrm{14}{x}=\mathrm{29}−\mathrm{29}=\mathrm{0} \\ $$$${x}^{\mathrm{2}} −\mathrm{7}{x}=\mathrm{0} \\ $$$$\Rightarrow{x}=\mathrm{0}\:{or}\:{x}=\mathrm{7} \\ $$$$\therefore{x}=\mathrm{7} \\ $$
Commented by mr W last updated on 11/Dec/25
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Answered by mr W last updated on 11/Dec/25
Commented by mr W last updated on 11/Dec/25
$$\left({x}−\mathrm{5}\right)+\left({x}−\mathrm{2}\right)=\mathrm{5}+\mathrm{2} \\ $$$$\Rightarrow{x}=\mathrm{5}+\mathrm{2}=\mathrm{7} \\ $$