a)Aplicam teorema catetei in ΔABC:
B.C=√EB*AB AC=√AE*AB
BC=√3*12 AC=√9*12
BC=√36 AC=√108
BC=6 cm AC=6√3 cm
ΔACB:C=90° si CE=inaltime
CE=BC*AC/AB
CE=6*6√3/12
CE=36√3/12
CE=3√3 cm
b)P ABCD=AB+BC+CD+AD
P ABCD=12+6+9+3√3
P ABCD=27+3√3
P ABCD=3(√3+9) cm
c)A ABCD=AD(AB+CD)/2
A ABCD=3√3*(12+9)/2
A ABCD=63√3/2 cm²