12)a)V(cub)=l³⇒l³=54√2⇒l=∛(54√2)=3√2cm→l=3√2cm
b)A(totala)=6·l²=6·(3√2)²=6·18=108cm²
c)m[∡(AD',B'C)]=m[∡(AD',DA')]=90°(diagonalele unui patrat sunt ⊥)
d)d(A,BD')→fie {O}∈BD';AO⊥BD'⇒AO=d(A,BD')
in ΔAOD':AD'=l√2=3√2·√2=6cm(formula diagonalei patratului)
BD'=l√3=2√2·√3=2√6cm(formula diagonalei cubului)
⇒inΔAOD' aplic th.pit.:AO²=AD'²-OD'²⇒AO=√(36-24)=√12=2√3cm
d(A,BD')=AO=2√3cm
13)a)A(totala)=6l²⇒6l²=96⇒l²=16⇒l=√16→l=4cm
b)V(cub)=l³=4³=64cm³
c)d(A',BD)→fie{O}=DB∩AC⇒A'O=d(A',BD)
in ΔA'AO aplic th.pit.:A'O²=A'A²+AO²⇒A'O=√(4²+(2√2)²=√(16+8=√24=2√6cm⇒d(A',BD)=2√6cm
AO=1/2·AC=1/2·l√2=4√2/2=2√2cm→(formula diag.patratului)
d)m[∡(B'C,B'B)=45°(deoarece diagonalele patratului se taie in ∡=90°
si se formeaza Δ isoscele cu ∡=45°)
