1.
a) BC = AD = 8 cm
Ducem DM si CN _I_ AB ( DM =CN)
AM = BN = (AB-CD) / 2 = (16-8) /2 = 4 cm
Δ ADM ≡ Δ NCB
In Δ ADM, dr. in M :
DM² = AD² -AM²
DM² = 64-16
DM² = 48
DM = h trapez = 4√3 cm
b) A = (B+b) × h / 2 = (AB+DC) × DM / 2 = (16+8) × 4√3 / 2 = 48√3 cm²
c) In Δ BDM , dr. in M :
BM =MN +BN = 8 +4 = 12 cm
BD² =BM² +DM²
BD² = 12² +(4√3)²
BD² = 192
BD = 8√3 cm
d) In Δ BDM, dr. in M :
DM = BD/2 = 8√3 / 2 = 4√3 (cateta este 1/2 din ipotenuza)
⇒ m(
