a. ducem DE⊥AB si CF⊥AB
ΔADE≡ΔCFB (triunghiuri dreptunghice, AD≡BC deoarece trapezul este isoscel si DE≡CF inaltimi in trapez), deci AE≡FB
BC=AE+EF+FB=2AE+DC, 18=2AE+12, 2AE=6, AE=FB=3
ΔADE Δdreptunghic, t lui Pitagora AD²=AE²+DE², DE²=AD²-AE²=81-9=72
DE=6√2=CF
A=(AB+CD)xDE/2=(18+12)x6√2/2=90√2
b. ΔACF Δdreptunghic AF=AE+EF=3+12=15
t lui Pitagora AC²=AF²+CF²=225+72=297, AC=3√33
