Salut ,
Problema I
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→→→ AO = 6 cm < = > OC = 6 cm
=> AC = 12 cm
AB = DC = 6√3
m( ‹ D ) = 90°
=> Teorema lui Pitagora
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AC² = AD² + DC² < = > 12² = AD² + ( 6√3 )² < = > 144 = AD² + 108 < = > AD² = 144 — 108 < = > AD² = 36 < = > AD = √36 = 6 cm
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→→→ A = L • l = 6√3 • 6 = 36√3 cm² — Aria ABCD
OC = OB = BC = 6 cm → ‹ OBC = 60°
‹ OBA = 90° — ‹ OBC = 90° — 60° = 30°
‹ MBS = ‹ OBA = 30° — ‹ opuse la vârf
=> BC = BM = 6 cm →
→→→ ‹ BCM = BMC = ( 180° — ‹ CBM ) : 2 = ( 180° — 90° — 30° ) : 2 = 60° : 2 = 30°
‹ OCM = OCB + BCM =
= 60° + 30° = 90°
→ BC² = AB • AS =>
→→→ 36 = 6√3 • BS =>
→→→ BS = 36 / 6√3 = 6 / √3= 6√3 / 3 = 2√3
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Problema II :
Teorema lui Pitagora in triunghiul ABC
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BC² = AB² + AC²
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BC = 20 m
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AM = AC • AB / BC
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AM = 9,6 m
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P = a + b + c =>
P = 12 + 16 + 20 =>
P = 48 m
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A = b • h / 2 < = >
A = 12 • 9,6 / 2 < = >
A = 115,2 / 2 < = >
A = 57,6 cm — Aria ABC
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MC = BC — MB =>
=> MC = 14 cm
=> MN = 4 cm
→→→ NC = 14 — 4
NC = 10 cm
A = 96 m² => 19,2 = x% • 96 m². < = > x% = 19,2 / 96 < = > x% = 0,2 m² →→ x = 20% — suprafața zambile
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