Răspuns:
Explicație pas cu pas:
BD=12√2 (diagonala patratului)
BO=6√2
In ΔVOB aplicam Pitagora
VB²=VO²+BO²
VO²=144-72=72
VO=6√2=OB⇒ΔVOB dreptunghic isoscel
VO=OC⇒ΔVOC dreptunghic isoscel⇒OM⊥VC
∡(MO,(VBD))=∡MOV=45°
b. ∡(VC,(ABC))=∡VCO=45°
c. ducem VN⊥BC si VP⊥AD ⇒ ∡((VBC),(VAD))=∡PVN
VN=VP=6√3 (inaltimi in Δechil)
PN=12
fie PE⊥VN
[tex]A_{VPN} =\frac{PN\cdot VO}{2}= 36\sqrt{2}=\frac{PE\cdot VN}{2} =\frac{PE\cdot 6\sqrt{3} }{2}[/tex]
PE=4√6
sin ∡PVN=PE:VP=4√6:6√3=[tex]\frac{2\sqrt{2} }{3}[/tex]
