Răspuns
Explicație pas cu pas:
a) Construim AP⊥BC (B intre C si P)
Daca MA⊥(ABC) si AP⊥BC (AP, BC ⊂(ABC)) , din T. 3 ⊥ => MP⊥BC
=> d(M, BC)=MP.
Calculam MP:
In ΔAPB (∡P=90°) mas(∡ABP)=mas(∡BAD)=60° -ca ∡-uri alterne
=> sin ∡ABP= AP/AB <=> √3/2=AP/16 => AP=16√3/2 => AP=8√3cm.
Daca MA⊥(ABC)=> MA⊥AP (AP⊂(ABC))
In ΔMAP(∡MAP=90°)-aplicam T. Pitagora :
MP²=MA²+AP²=(6√3)²+(8√3)²=36 3+64 3=100 3 => MP=√300=10√3
=> d(M, BC)=10√3cm.
b) Construim AR⊥DC (D intre R si C)
Daca MA⊥(ABC) si AR⊥DC (AR, DC ⊂(ABC)) , din T. 3 ⊥ => MR⊥DC
=> d(M, DC)=MR.
∡ADR≡∡DAB ( alterne interne) => mas∡(ADR)=60°
In ΔADR aplicam sinus: sin∡ADR=AR/AD sin 60°=AR/12 => √3/2=AR/12
=> AR=6√3cm.
In ΔMAR(∡MAR=90°)-aplicam T. Pitagora :
=> MR²=MA²+AR²=(6√3)²+(6√3)²=3 ·6²·2 => MR=6√6cm.
c) d(M, (MBC))=0 deoarece M⊂MBC.
