Răspuns:
Explicație pas cu pas:
AB=AO√3, deci AO=AB/√3=12√3/√3=12m, A'O'=A'B'/√3=4√3/√3=4m.
Trasam A'K⊥AO, K∈AO, atunci A'K=OO'=6m. AK=AO-A'O'=12-4=8m.
Din ΔA'AK, ⇒A'A²=A'K²+AK²=6²+8²=36+84=100, deci A'A=√100=10m
Atunci LungimeaProfil=4·A'A=4·10=40m.
b) Aria(laterala)=3·Aria(BCC'B'), unde BCC'B' este trapez isoscel.
Trasam C'M⊥BC, M∈BC, atunci CM=(AB-A'B')/2=4√3m.
⇒C'M²=C'C²-CM²=10²-(4√3)=100-48=52=4·13, deci C'M=2√13m apotema sau inaltimea trapezului. Atunci Aria(laterala)=3·Aria(BCC'B')=3·C'M·(BC+B'C')/2=3·2√13·(12√3+4√3)/2= 3√13·16√3=48√39m².
c) ∡(CC',BB')=∡(DC,DB).
ΔDOE≅ΔDO'E', fie DO=H, iar DO'=H-h=H-6. EO=(1/2)·AO=6m, E'O'=(1/2)·A'O'=2m. Din asemanare, ⇒DO/D'O'=EO/E'O', ⇒H/(H-6)=6/2=3, deci H=3·(H-6), ⇒H=3H-18, ⇒H-3H=-18, ⇒-2H=-18, ⇒H=9m=DO.
Din ΔDOE, ⇒DE²=DO²+EO²=9²+6²=3²·3²+3²·2²=3²·13, deci DE=3√13m.
⇒Aria(ΔDBC)=(1/2)·BC·DE=(1/2)·12√3·3√13=18√39m²
Dar Aria(ΔDBC)=(1/2)·DC·DB·sin(∡CDB)
Din ΔDCE, ⇒DC²=DE²+CE²=(3√13)²+(6√3)²=9·13+36·3=9·(13+12)=9·25
Deci DC=√(9·25)=3·5=15m.
⇒(1/2)·DC·DB·sin(∡CDB)=18√39, ⇒(1/2)·15·15·sin(∡CDB)=18√39 |·2, ⇒
15·15·sin(∡CDB)=36√39, ⇒sin(∡CDB)=36√39:(15·15)=4√39/25=0,16√39.
