Question

imi trebuie rezolvari la problemele de la 3 la 7​

Answer 1

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

Answer 2

Răspuns:

Explicație pas cu pas:

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 par.