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Distanțe în cub
Fie cubul ABCDA’B’C’D’ de latură a.
  Aflați:
  a) d(A’; D);
  b) d(A;BB’);d(A’;CC’);d(C;AC’);d(A’;BD);
  c) d(D’; (ABC)); d(A; (A’BD));
  d) d(A’D’;AD); d(A’D’;BC); d(A’D; B’C);
  e) d(A’B’; (DCC’));
  f) d((A’AD); (B’C’C)).
1.a)




       d(A’; D) = ?
1.a)




d(A’; D) = A’D =   a 2 + a 2 = 2a 2 = a 2
1.b1)




        d(A; BB’) = ?
1.b1)




AB ⊥ BB’ => d(A; BB’) = AB = a
1.b2)




        d(A’; CC’) = ?
1.b2)




CC’ ⊥ (A’B’C’), A’C’ ⊂ (A’B’C’) => CC’ ⊥ A’C’


             d(A’; CC’) = A’C’
1.b2)




∆A’B’C’, (m<(B’) = 90o) => A’C’ = a 2


        d(A’; CC’) = A’C’ = a 2
1.b3)




        d(C; AC’) = ?
1.b3)




   CE ⊥ AC’ => d(C; AC’) = CE
1.b3)




  CE = h∆ACC’; ∆ACC’ = dreptunghic

                                     a 6
d(C; AC’) = CE = (AC . CC’)/AC’ =
                                      3
1.b4)




        d(A’; BD) = ?
1.b4)




   A’O ⊥ DB => d(A’; DB) = A’O
1.b4)




DB=A’D=A’B= a 2 => ∆A’BD = echilateral

                                 a 6
     d(A’;BD) = A’O = h∆A’BD =
                                  2
1.c1)




        d(D’; (ABC)) = ?
1.c1)




D’D ⊥ (ABC) => d(D’; (ABC)) = D’D = a
1.c2)




        d(A; (A’BD)) = ?
1.c2)




AH ⊥ (A’BD) => d(A; (A’BD)) = AH
1.c2)




    AH = h∆A’AO; ∆A’AO = ∆ dreptunghic

                                         a 3
d(A; (A’BD)) = AH = (AO . AA’)/OA’ =
                                          3
1.d1)




        d(A’D’; AD) = ?
1.d1)




 A’D’ | | AD; AA’ ⊥ AD; AA’ ⊥ A’D’


        d(A’D’; AD) = AA’ = a
1.d2)




        d(A’D’; BC) = ?
1.d2)




 A’D’ | | BC; A’B ⊥ BC; A’B ⊥ A’D’


   d(A’D’; BC) = A’B = a 2
1.d3)




        d(A’D; B’C) = ?
1.d3)




 A’D | | B’C; A’B’ ⊥ A’D; A’B’ ⊥ B’C


    d(A’D; B’C) = A’B’ = a
1.e)




       d(A’B’; (DCC’)) = ?
1.e)




A’B’ | | (DCC’); B’C’ ⊥ A’B’; B’C’ ⊥ (DCC’)


     d(A’B’; (DCC’) = B’C’ = a
1.f)




       d((A’AD); (B’C’C)) = ?
1.f)




(A’AD) | | (B’C’C); AB ⊥ (A’AD); AB ⊥ (B’C’C)


      d(A’AD); (B’C’C)) = AB = a

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Distanțe în cub sub 40 caractere

  • 1. Distanțe în cub Fie cubul ABCDA’B’C’D’ de latură a. Aflați: a) d(A’; D); b) d(A;BB’);d(A’;CC’);d(C;AC’);d(A’;BD); c) d(D’; (ABC)); d(A; (A’BD)); d) d(A’D’;AD); d(A’D’;BC); d(A’D; B’C); e) d(A’B’; (DCC’)); f) d((A’AD); (B’C’C)).
  • 2. 1.a) d(A’; D) = ?
  • 3. 1.a) d(A’; D) = A’D = a 2 + a 2 = 2a 2 = a 2
  • 4. 1.b1) d(A; BB’) = ?
  • 5. 1.b1) AB ⊥ BB’ => d(A; BB’) = AB = a
  • 6. 1.b2) d(A’; CC’) = ?
  • 7. 1.b2) CC’ ⊥ (A’B’C’), A’C’ ⊂ (A’B’C’) => CC’ ⊥ A’C’ d(A’; CC’) = A’C’
  • 8. 1.b2) ∆A’B’C’, (m<(B’) = 90o) => A’C’ = a 2 d(A’; CC’) = A’C’ = a 2
  • 9. 1.b3) d(C; AC’) = ?
  • 10. 1.b3) CE ⊥ AC’ => d(C; AC’) = CE
  • 11. 1.b3) CE = h∆ACC’; ∆ACC’ = dreptunghic a 6 d(C; AC’) = CE = (AC . CC’)/AC’ = 3
  • 12. 1.b4) d(A’; BD) = ?
  • 13. 1.b4) A’O ⊥ DB => d(A’; DB) = A’O
  • 14. 1.b4) DB=A’D=A’B= a 2 => ∆A’BD = echilateral a 6 d(A’;BD) = A’O = h∆A’BD = 2
  • 15. 1.c1) d(D’; (ABC)) = ?
  • 16. 1.c1) D’D ⊥ (ABC) => d(D’; (ABC)) = D’D = a
  • 17. 1.c2) d(A; (A’BD)) = ?
  • 18. 1.c2) AH ⊥ (A’BD) => d(A; (A’BD)) = AH
  • 19. 1.c2) AH = h∆A’AO; ∆A’AO = ∆ dreptunghic a 3 d(A; (A’BD)) = AH = (AO . AA’)/OA’ = 3
  • 20. 1.d1) d(A’D’; AD) = ?
  • 21. 1.d1) A’D’ | | AD; AA’ ⊥ AD; AA’ ⊥ A’D’ d(A’D’; AD) = AA’ = a
  • 22. 1.d2) d(A’D’; BC) = ?
  • 23. 1.d2) A’D’ | | BC; A’B ⊥ BC; A’B ⊥ A’D’ d(A’D’; BC) = A’B = a 2
  • 24. 1.d3) d(A’D; B’C) = ?
  • 25. 1.d3) A’D | | B’C; A’B’ ⊥ A’D; A’B’ ⊥ B’C d(A’D; B’C) = A’B’ = a
  • 26. 1.e) d(A’B’; (DCC’)) = ?
  • 27. 1.e) A’B’ | | (DCC’); B’C’ ⊥ A’B’; B’C’ ⊥ (DCC’) d(A’B’; (DCC’) = B’C’ = a
  • 28. 1.f) d((A’AD); (B’C’C)) = ?
  • 29. 1.f) (A’AD) | | (B’C’C); AB ⊥ (A’AD); AB ⊥ (B’C’C) d(A’AD); (B’C’C)) = AB = a