3)
∡C = 45° => ΔABC = isoscel => AB = AC = 48/√2 = 48√2/2 = 24√2 cm
A = AB·AC/2 = 24·24·2/2 = 576 cm²
4)
CD = 10 cm
2CD = AB } => AB = 20 cm
DUc CE ⊥ AB
∡B = 45° => ΔCEB = dr. isoscel => CE = EB = 10 cm
BC = CE√2 = 10 √2 cm
P = AB+BC+CD+AD
= 20cm + 10√2cm + 10cm + 10cm
= (40+10√2) cm
A = (AB+CD/2)×AD = (10+20)/2×10 = 15×10 = 150 cm²
5)
∡NMP = 30° => ∡NMQ = 2×30° = 60°
(diagonala e bisectoare)
∡NMQ = 60° => ΔNMQ = echilateral => MN = 10 cm
P = 4MN = 40 cm
A(MNPQ) = 2A(MNQ) = 2×100√3/4 = 50√3 cm²
6)
A(ABCD) = AC×BD/2 = 18×24/2 = 216 cm²
Duc BE ⊥ AD
BE×AD/2 = A(ABCD)/2
BE×AD = 216
BE = 216/AD
AD = √(AC/2)²+(BD/2)² = √144+81 = √225 = 15 cm
BE = 216/15 = 14,4 cm
7)
∡G = ∡E => ∡AEF = ∡HGB (unghiuri date de bisectoare) (1)
AG || BE (2)
Din (1) si (2) => AGBE = paralelogram
