5.
BD=48cm
[tex]BO= \frac{BD}{2}=24 [/tex]
[tex]In [/tex] Δ [tex]BOC(teorema lui Pitagora)=\ \textgreater \ [/tex]
[tex] BC^{2} = BO^{2} +CO^{2} [/tex]
[tex]CO ^{2} = BC^{2} - BO^{2} [/tex]
[tex] CO^{2} =676-576[/tex]
[tex]CO^{2} =100[/tex]
[tex]CO=10cm[/tex]
[tex]AC=CO*2=20cm[/tex]
[tex]A_{ABCD} = \frac{ d_{1}* d_{2} }{2} = \frac{BD*AC}{2} = \frac{48*20}{2} =480 cm^{2} [/tex]
6.
[tex]DC=8cm[/tex]
[tex]DA=CM=9cm[/tex]
Δ[tex]CMB(dreptunghic)=CB=15cm[/tex]
[tex]In[/tex] Δ [tex]BMC(teorema lui Pitagora)=\ \textgreater \ [/tex]
[tex] CB^{2} = CM^{2} + MB^{2} [/tex]
[tex] MB^{2} = CB^{2} - CM^{2} [/tex]
[tex] MB^{2} = 225-81[/tex]
[tex] MB^{2} =144[/tex]
[tex]MB=12cm[/tex]
[tex]AB=AM+MB[/tex]
[tex]AB=8+12[/tex]
[tex]AB=20cm[/tex]
[tex]A_{ABCD} = \frac{(b+B)*h}{2}= \frac{(AB+DC)*CM}{2} = \frac{28*9}{2} =14*9=126cm^{2} [/tex]
O zi buna! Sper ca ti-am fost de ajutor! :)
