Răspuns :
AB=12√2
VA=8√6
a.
Fie AD⊥BC
- AD este inaltime intr-un triunghi echilateral
[tex]AD=\frac{l\sqrt{3} }{2}[/tex]
[tex]AD=\frac{12\sqrt{6} }{2} =6\sqrt{6}[/tex]
OA se afla la 2 treimi de baza
[tex]OA=\frac{2}{3} \times \ 6\sqrt{6} =4\sqrt{6}[/tex]
OA=4√6cm
b. AD⊥BC
AD⊂(ABC)
∡(VA,(ABC))=∡VAD=∡VAO
AD=6√6
Fie VO⊥AD
- Aplicam Pitagora in ΔVAO dr in O
VA²=VO²+OA²
VO²=384-96
VO²=288
VO=12√2cm
Observam ca 2AO=VA⇒ conform reciprocei unghiului de 30°⇒ m(∡AVO)=30°⇒ m(∡VAO)=60°
c.
d(O,(VBC))=OE
VD⊥BC
OD⊥BC
BC⊂(VBC)
OD se afla la o treime din AD
[tex]OD=\frac{1}{3} \times 6\sqrt{6} =2\sqrt{6}[/tex]
OD=2√6cm
In ΔVOD aplicam Pitagora sa aflam
VD²=VO²+OD²
VD²=288+24=312
VD=√312=2√78
OE inaltime in ΔVOD
[tex]OE=\frac{VO\times OD}{VD} =\frac{12\sqrt{2}\times 2\sqrt{6} }{2\sqrt{78} }=\frac{12\sqrt{6} }{\sqrt{39} } =\frac{4\sqrt{234} }{13}[/tex]
