a.
AB=6cm
Diagonala unui cub=l√3
Deci BH=AG=6√3cm
b.
HC=l√2 (diagonala patrat)
HC=6√2cm
BH²=HC²+BC²
108=72+36
108=108⇒ R.T.P⇒ ΔBHC dr in C
[tex]d(C,BH)=\frac{CH\times BC}{BH} \ (formula\ inaltimii\ in\ triunghiul\ dreptunghic)\\\\d(C,BH)=\frac{6\sqrt{2}\times 6 }{6\sqrt{3} } =\frac{6\sqrt{6} }{3} =2\sqrt{6} \ cm[/tex]
c.
AB⊥BC
CM⊥BG⇒ d(A,(BCG)=AM
- In ΔAMB dr in M aplicam Pitagora
AM²=AB²+BM²
BM=6√2:2=3√2 cm
AM²=36+18
AM²=54
AM=3√6 cm
d.
DF=6√3 cm(diagonala cub)
ΔDFB dr in B
Fie BN⊥FD
BN=2√6 cm=d(C,BH)
EF⊥BF
BN⊥FD⇒ d(E,(BDF))=EN
EF²=36
DE²=72
DF²=108⇒DF²=EF²+DE²⇒ R.T.P ⇒ΔEDF dr in E⇒ EN inaltime in triunghi
[tex]EN=\frac{ED\times EF}{FD} =\frac{6\sqrt{2} \times 6}{6\sqrt{3} } =2\sqrt{6}\ cm[/tex]
