Avem ΔAHC care este isoscel ∠HCA≡∠HAC=45⁰ ⇒ AH≡HC
ΔAHC avem th Pitagora AC²=AH²+HC²=2AH²
16²=2AH²
AH=256/2=128=64x2=8√2
AH=8√2
PENTRU AFLAREA PERIMETRULUI AVEM NEVOIE DE DIMENSIUNEA LATURILOR
In ΔABH aplicam th ∠30⁰⇒AH=AB/2⇒AB=2AH=16√2
th Pitagora AB²=AH²+BH²⇒BH²=AB²-AH²=512+128=640
BH=8√10
BC=BH+HC=8√10+8√2
P=AB+BC+CA=16√2+8√10+8√2+16=16+24√2+8√10=16+33.94+25.29
P=75.23 cm
