[tex]Se \;da: \\ AB = 12\;dm \\ BC=18\;dm \\ P\;este\;mijlocul\;lui\;CD\;\; iar\;\;M\;este\;mijlocul\;lui\;AB \\ \\ Rezolvare: \\ \\ a) \\ A_{ABCD} = AB \;\cdot\;BC = 12 \;\cdot\;18 = \boxed{216 \;dm^2} = \boxed{2,16 \;m^2} \\ \\ b) \\ BC=AD \; fiind\; laturi\;opuse \;ale \;dreptunghiului. \\ CP = PD \; deoarece P \; este\;mijlocul\;segmentului\;CD. \\ \ \textless \ ADP = \ \textless \ BCP = 90^o BCP [/tex]
[tex]\\ =\ \textgreater \ \;\Delta ADP \equiv \Delta BCP \\ =\ \textgreater \ \; [AP] \equiv [BP] \\ \Delta APB \;este\;isoscel. \\ AP = BP = \sqrt{BC^2+ (\frac{CD}{2})^2 } =\sqrt{18^2+ (\frac{12}{2})^2 }= \\ \\ =\sqrt{18^2+ 6^2 }=\sqrt{324+ 36 }=\sqrt{360 }=6 \sqrt{10}\;dm \\ AB=12\;dm \\ P_{ABP}=AB+AP+BP= \\ =12+6 \sqrt{10}+6 \sqrt{10} = 12+12\sqrt{10} = \boxed{12(1+ \sqrt{10})\;dm}[/tex]
[tex]c) \\ In \;triunghiurile \Delta MNB \;si\; \Delta DNC \;avem: \\ \ \textless \ MBN \; = \; \ \textless \ DCN = 90^o \\ \ \textless \ MNB = \ \textless \ CND \\ =\ \textgreater \ \; \Delta MNB \;si\; \Delta DNC \;\;sunt\;triunghiuri\;dreptunghice\; si\; asemenea. \\ \frac{MB}{CD}= \frac{BN}{CN} = \frac{1}{2} \;\;\;\; =\ \textgreater \ \;CN =2BN \\ =\ \textgreater \ \; BC=3BN =\ \textgreater \ BN = \frac{18}{3} = = 6\;dm \\ MB = \frac{AB}{2} = \frac{12}{2} = 6\;dm \\ =\ \textgreater \ \;MB=BN \\ =\ \textgreater \ \; \Delta MNB \;\;si \;\; \Delta DNC \;\;\; sunt\; triunghiuri \;dreptunghice isoscele. \\ =\ \textgreater \ \;\;\ \textless \ MNB = \ \textless \ CND = 45^o [/tex]
[tex]=\ \textgreater \ \;\;\ \textless \ MND = 180^o - 45^o-45^o =\boxed{ 90^o} \\ =\ \textgreater \ \;\;\boxed{MN\; \bot \;ND}[/tex]
