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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

$$set{B}^{\prime }M=//BN$$$$set{D}^{\prime }M=//DQ$$$$\mathrm{\Delta }{ABB}^{\prime }\equiv \mathrm{\Delta }{ADD}^{\prime }$$$$\Rightarrow {AB}^{\prime }={AD}^{\prime }and{AB}^{\prime }\mathrm{\perp }{AD}^{\prime }$$$$...$$$$\Rightarrow {AB}^{\prime }{C}^{\prime }{D}^{\prime }issquare.$$$${AM}^2+C^{\prime }{M}^2=B^{\prime }{M}^2+D^{\prime }{M}^2$$$$\Rightarrow x^2+2^2=6^2+7^2$$$$\Rightarrow AM=x=\sqrt{6^2+7^2-2^2}=9✓$$

Commented by mnjuly1970 last updated on 12/Oct/23

$$verynicesolution$$$$thanksalotsirW$$

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