a) At = Al + 2 Ab
Al= Pb ×h
V= Ab ×h
Ab = l² √3/4 = 16√3/4 = 4√3 cm²
Al = 12 ×4√3 = 48√3 cm²
At = 48√3 + 2× 4√3 = 48√3 +8√3 = 56√3 cm²
V= 4√3 × 4√3 = 48 cm³
b) In Δ FCC` (dreptunghic in C) , FC = h in Δ ABC
(FC`)² = FC² + (CC`)²
FC = l√3/2 = 4√3/2 = 2√3
(FC`)²= (2√3)² + (4√3)²
(FC`)²= 12 + 48
(FC`)²= 60
distanta de la C` la AB = FC` = 2√15 cm