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(<MBD) = 30 grade ( cf.teoremei unghiului de 30 grade)
si
m(<ABD) = m(<MBD) =
30 grade⇒m(<MDB) = 60 grade
In Δ ADM, dr. in M :
cateta AM este 1/2 din ipotenuza AD
AM = AD/2 = 8/2 = 4
⇒ m(<ADM) = 30 grade
m(<ADB) = m(<ADM) + m(MDB) = 30 + 60 =
90 grade
2. In trapezul dreptunghic ortodiagonal:
h = √ (B ×b) (inaltimea este egala cu media geometrica a celor 2 baze)
⇒ h=√(20×5) =√100 = 10 cm
A = (B+b) × h / 2 = 25 × 10 / 2 = 125 cm²
Notam trapezul ABCD, AB II CD, m(<DAB)=m(<ADC) = 90 grade
h= AD = CN = 10 cm (CN _I_AB)
In Δ CNB, dr. in N :
NB = AB - CD = 20 -5 = 15 cm
BC² = CN² +NB²
BC² = 10² + 15²
BC² = 325
BC = 5√13 cm
P trapez ABCD = AB +BC +CD +AD = 20 +5√13 +5 +10 =35 + 5√13 =5(7+√13) cm
