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Konštrukcia lichobežníka

Konštrukcia lichobežníka. Autor: Mgr. Ľubica Nemčovičová. D: z 1 =6,5 cm z 2 =3,5 cm b =5 cm e =7 cm Z: lichobežník ABCD. Rozbor: Poznáme: A,B Hľadáme: C,D Podmienky pre C: k 1 ; k 1 (A; 7cm) k 2 ; k 2 (B; 5cm) Ck 1 k 2 Podmienky pre D: p; p ║AB, Cp

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Konštrukcia lichobežníka

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  1. Konštrukcia lichobežníka Autor: Mgr. Ľubica Nemčovičová

  2. D: z1=6,5 cm z2=3,5 cm b =5 cm e =7 cm Z: lichobežník ABCD Rozbor: Poznáme: A,B Hľadáme: C,D Podmienky pre C: k1; k1(A; 7cm) k2; k2(B; 5cm) Ck1k2 Podmienky pre D: p; p║AB, Cp k3; k3(C; 3,5cm) Dpk3 k1 k2 3,5 cm D C p k3 5 cm 7 cm 6,5 cm B A Lichobežník

  3. Postup: AB;AB=6,5cm k1; k1(A; 7cm) k2; k2(B; 5cm) C; Ck1k2 p; p║AB, Cp k3; k3(C; 3,5cm) D; Dpk3 lich. ABCD Konštrukcia: k3 k1 k2 C D p A B

  4. D: a= 7cm b= 5cm c= 3cm d= 4,5cm Z: lich. ABCD Rozbor: Poznáme: A,B Hľadáme: C,D Podmienky pre C: k1; k1(B; 5cm) k2; k2(X; 4,5 cm) Ck1k2 Podmienky pre D: (vaša tvorba ) XBC vieme zostrojiť(sss) AXCD je kosodĺžnik k1 3 cm D C k2 5 cm 4,5 cm 7 cm X B A Lichobežník 2 4,5 cm (7 – 3)cm 4 cm

  5. RIEŠ SAMOSTATNE: D: a= 5cm b= 4cm = 60° = 75° Z: lich. ABCD Y X C D p 4 cm   A B 5 cm sus

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