VAC triunghi echilateral
b.
VA=AC=VC=12cm
AO=12:2=6cm
Aplicam Pitagora in ΔVAO dr in O
VA²=VO²+OA²
VO²=144-36=108
VO=6√3cm
c.
AC este diagonala in patratul ABCD,
d=l√2
Adica 12=AB√2
AB=6√2cm
d.Fie VM⊥AB
VM⊂(VAB)
OM⊥AB
d(O,(VAB))=ON
OM=6√2:2=3√2 (apotema bazei)
Aflam VM din Pitagora in ΔVOM
VM²=OM²+VO²
VM²=18+108=126
VM=√126
ON este inaltime in ΔVOM
[tex]ON=\frac{VO\times OM}{VM} =\frac{6\sqrt{3}\times 3\sqrt{2} }{\sqrt{126} } =\frac{18}{\sqrt{21} } =\frac{6\sqrt{21} }{7}[/tex]
