a) BC² = AC² -AB² = 64 - 48 = 16 BC = 4cm
V = Ab·h h = 32√3/16√3 = 2cm
Al = Pb·h = 2(4√3 + 4)·2 = (16√3 + 16)cm²
At = Al +2Ab = 16√3 +16 + 16√3 = 32√3 +16
b) in Δ B'AC isoscel (B'A =B'C ) d(B',AC) = B'O ( BD∧AC ={O} )
in ΔB'BO (dreptunghic) BB' = 2 BO =4 ⇒ B'O = √(4+16) = 2√5
d[B,(B'AC)] = d(B,OB') din ΔBB'O
daca BE = inaltime in Δ B'BO BE² = m·n (m = B'E n = EO)
BE² = B'B² - m² BE² = BO²- n² 4-m² = 16 - n² n² -m² =12 (n-m)(n+m) =12
2√5(n-m) = 12 n-m = 6√5/5 n+m = 2√5 2n = 16√5/5 n = 8√5/5 m= 2√5 -8√5/5 =
= 2√5/5 ⇒ BE² = 2√5/5 ·8√5/5 = 16/5 BE = 4√5/5cm
< (B'AC) ,(BAC) = <B'OM (OM perpendicular pe AB si = BC/2 = 2cm)
in Δ B'MO B'M = BB' = 2cm MO = 2cm ⇒ Δ B'MO = dreptunghic isoscel ⇒ mas< =45gr.
