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Probleme. Ipoteza: ABC-∆ M€(BC) MP||AB; MN||AC Concluzia: 1 Demonstratie:. A P N

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Probleme

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  1. Probleme Ipoteza: ABC-∆ M€(BC) MP||AB; MN||AC Concluzia: 1 Demonstratie: A P N B C MN AB MP AC Tf~ ∆CAB (MN||AB) => Tf~ ∆BAC (MP||AC) => 1 CN CA CM CB MN AB BP BA BM BC MP AC MN AB MP AC CM CB BM BC CM+BM BC BC BC

  2. A 18 24 M N 12 16 B C 50 Ipoteza: ABC-∆ AB=30 cm; AC=40 cm; BC=50 cm M€(AB); AM=18 cm N€(AC); CN=16 cm P€(BC); NP||AB Concluzia: a) MN||BC b) P patrulaterului BMNP Demonstratie: • MB=AB-AM=30-18=12 cm • AN=AC-NC=40-16=24 cm • b) MN||BC => MN||BP • NP||AB => NP||MB BMNP paralelogram • P BMNP = 2(MN+MB) • Tf~ • ∆ABC: MN||BC => MN= 30 • P BMNP = 2(30+12)= 2X42= 84 cm AM MB 18 12 3 2 R.T.Th MN||BC AM MB AN NC AN NC 24 16 3 2 AM AB AN AC MN BC 18 30 MN 50 18X50 30

  3. V 10,05 T 3,35 1,5 R S 9 B 3 Ipoteza: SV=12 m TB||VR; TB=1.5 m BS=3 m TS=3.35 m Concluzia: RV+VS=? Demonstratie: RB=RS-BS=12-3=9 m ∆VRS (TB||VR) =>  => TV=10,05 m Tf~ ∆SRV (TB||VR) => SV= = 13,40 m VR= = 6 m h=VR+SV=6+13.40-19.40 m ST TV SB BR 3.35 TV 3 9 ST SN SB SR TB VR 3,35X12 3 12X1,5 3

  4. V A 0.75 O 0.5 B D C E Ipoteza: BC=1.5 m OB= 0.5 m AB = 75 cm OD=15 m Concluzia: VE=? h=VD+DE Fie DE┴sol => BCED dreptunghi => => DE=BC=1.5 m Tf~ ∆ODV (AB||VD) => => => OA OV OB OD AB VD OA OV 0.5 15 0.75 VD 15x0.75 0.5 75 100 10 5 45 2 VD= 15x x 22.5 m h=VD+DE= 1.5+22.5= 24m

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