1. mas<AOB/7 = mas<AOC/6 = mas<BOC/5 = 180/18 = 10
mas < AOB = 7·10 = 70
mas < AOC = 6·10 = 60
mas <BOC = 5 ·10 = 50
2. daca mas < A = 90, AM = mediana (BM = MC) si AD = inaltime (AD_|_BC)
AM = BC/2 ⇒ BC = 2·5 = 10 cm
... in Δ ABC = BC·AD/2 = 10·4/2 = 20cm²