Question

ex 7 si dau si coroana

Answer 1

   
[tex]\displaystyle \\ MA = MB ~~~si~~~ QD=QC\\ \Longrightarrow~~ MQ ~\text{ este linie mijlocie in trapezul }~ABCD \\ NB=ND~~~si~~~PA=PC \\ \Longrightarrow~~ N = BD \cap MQ ~~~si~~~ P = AC \cap MQ \\ \Longrightarrow~~ M, N,P,Q ~\text{ sunt coliniare} \\ MN~\text{ este linie mijlocie in }\Delta ABD \\ PQ~\text{ este linie mijlocie in }\Delta ACD \\ MP~\text{ este linie mijlocie in }\Delta ABC \\ NQ~\text{ este linie mijlocie in }\Delta BDC [/tex]


[tex]\displaystyle \\ \text{Rezolvare: } \\ \\ a) AD=4~cm ~~~si~~~BC=10~cm\\ \\ MN= \frac{AD}{2} =\frac{4}{2} = \boxed{2~cm }\\ \\ MP= \frac{BC}{2} =\frac{10}{2} = \boxed{5~cm } \\ \\ MQ= \frac{AD+BC}{2} =\frac{4+10}{2} =\frac{14}{2} = \boxed{7~cm }\\ \\ NP= MP - MN = 5 - 2=\boxed{3~cm }[/tex]


[tex]\displaystyle \\ b) AD=6~cm ~~~si~~~BC=9~cm\\ \\ NQ= \frac{BC}{2} = \frac{9}{2} = \boxed{4,5~cm}\\ MQ= \frac{AD+BC}{2} = \frac{6+9}{2} = \frac{15}{2} =\boxed{7,5~cm}\\ PQ= \frac{AD}{2} = \frac{6}{2} = \boxed{3~ cm}\\ NP= NQ-PQ = 4,5 - 3 = \boxed{1,5~cm} [/tex]