Question Number 198158 by mnjuly1970 last updated on 12/Oct/23
Answered by mr W last updated on 12/Oct/23
Commented by mr W last updated on 12/Oct/23
$${set}\:{B}'{M}=//{BN} \\ $$$${set}\:{D}'{M}=//{DQ} \\ $$$$\Delta{ABB}'\equiv\Delta{ADD}' \\ $$$$\Rightarrow{AB}'={AD}'\:{and}\:{AB}'\bot{AD}' \\ $$$$… \\ $$$$\Rightarrow{AB}'{C}'{D}'\:{is}\:{square}. \\ $$$${AM}^{\mathrm{2}} +{C}'{M}^{\mathrm{2}} ={B}'{M}^{\mathrm{2}} +{D}'{M}^{\mathrm{2}} \\ $$$$\Rightarrow{x}^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} =\mathrm{6}^{\mathrm{2}} +\mathrm{7}^{\mathrm{2}} \\ $$$$\Rightarrow{AM}={x}=\sqrt{\mathrm{6}^{\mathrm{2}} +\mathrm{7}^{\mathrm{2}} −\mathrm{2}^{\mathrm{2}} }=\mathrm{9}\:\checkmark \\ $$
Commented by mnjuly1970 last updated on 12/Oct/23
$$\:\:{very}\:{nice}\:{solution} \\ $$$${thanks}\:{alot}\:\:{sir}\:\:{W} \\ $$