Question

Ex 2, va rog mult.....

Answer 1

a) AM =inaltime in tr.echilateralde latura AB=4√2
deci AM=AB√3/2= 4√2*√3/2=2√6


b)vol tetraedruluiregulat = Ab*inaltimea /3=(1/3)* AB²[(√3)/4]*AB√2/√3=AB³√2/12=
(4√2)³ *√2/12=64*2√2*√2/12=64*2*2/12=64/3 cm³
am folosit inaltimeatetraedruluireg = AB*√2/√3


c)fie MP||AB. p∈AC
m∡(AB, MN)=m∡(MP,MN)
 MP=AB/2 ca l.m.inΔABC
analog NP=AB/2 l.m inΔACD
deci MPN isoscel de baza MN


AM⊥BC, DM⊥BC⇒(AMC)⊥BC, BC⊥(AMC)⇒BC⊥MN⊂(AMC)⇔MNCdreptunghic in M

MN= √((AB√3/2)²-(AB/2²)=AB√2/2

MP=NP=AB/2

din cele 2 relde mai sus⇒MN²=NP²+MP²⇒(Tec T.Pitagora) ΔMNP dreptunghic in P
cumΔMNP e si isocel⇒m∡(MP,MN)=45°=m∡(AB, MN)