a.
∡(VA,(ABC))=∡VAC=60°
AB=6 cm
VA=VC (muchii piramida)
∡VAC=60°⇒ ΔVAC echilateral
VA=VC=AC
AC=l√2
AC=6√2 cm
VA=6√2 cm
b.
Fie VM apotema piramidei
Aplicam Pitagora in ΔVMB
VB²=VM²+BM²
72=VM²+9
VM=3√7 cm
Fie ON⊥VM,
trebuie sa demonstram ca N este ortocentrul ΔVBC
OM=3 cm
In ΔVOM
VO²+OM²=VM²
63=VO²+9
VO=3√6 cm
OM²=MN×VM
9=MN×3√7
[tex]MN=\frac{3\sqrt{7} }{7}[/tex]
Fie BT⊥VC si E=VM∩BT, E ortocentrul VBC
VM×BC=BT×VC
3√7×6=BT×6√2
[tex]BT=\frac{3\sqrt{14} }{2}[/tex]
In ΔBTC, aplicam Pitagora
BC²=TC²+BT²
[tex]TC^2=36-\frac{126}{4} =\frac{18}{4}\\\\ TC=\frac{3\sqrt{2} }{2}[/tex]
ΔBTC~ΔBME
[tex]\frac{BT}{BM}=\frac{BC}{BE} =\frac{TC}{ME}[/tex]
[tex]\frac{\frac{3\sqrt{14} }{2} }{3} =\frac{\frac{3\sqrt{2} }{2} }{EM}[/tex]
[tex]EM=\frac{9\sqrt{2} }{2} \times \frac{2}{3\sqrt{14} } =\frac{3\sqrt{7} }{7}[/tex]
MN=EM⇒ E si N coincid⇒ N ortocentru
