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waheguru Group Title

by connecting the midpoints of a reatangle u get a rhombus right?

  • 2 years ago
  • 2 years ago

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  1. FoolForMath Group Title
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    yes.

    • 2 years ago
  2. jimmyrep Group Title
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    yup

    • 2 years ago
  3. Jemurray3 Group Title
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    In general you get a parallelogram, which does not necessarily have to be a rhombus.

    • 2 years ago
  4. FoolForMath Group Title
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    No you will get a rhombus always.

    • 2 years ago
  5. FoolForMath Group Title
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    and square gives square.

    • 2 years ago
  6. Jemurray3 Group Title
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    Oh my, I was thinking of something else, I apologize. Yep :)

    • 2 years ago
  7. Jemurray3 Group Title
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    A square is a rhombus though.

    • 2 years ago
  8. FoolForMath Group Title
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    No problem :)

    • 2 years ago
  9. FoolForMath Group Title
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    If you join the midpoints of a square the secondary figure is also a square.

    • 2 years ago
  10. Jemurray3 Group Title
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    Yes, but I meant that a square is also a rhombus.

    • 2 years ago
  11. FoolForMath Group Title
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    Every square is rhombus, rectangle and parallelogram.

    • 2 years ago
  12. Jemurray3 Group Title
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    I disagree. A rhombus is a quadrilateral with all four sides being of the same length. A square is an example of this.

    • 2 years ago
  13. Jemurray3 Group Title
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    I don't mean that to be a square is to be a rhombus, I mean that a square happens to be a particular example of a rhombus.

    • 2 years ago
  14. FoolForMath Group Title
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    Yes it's always precise to use the subset instead of super-set when your conditions satisfies the subset. Say what is 2 ? It is an integer and also real but we generally say integer.

    • 2 years ago
  15. Jemurray3 Group Title
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    Why not natural then?

    • 2 years ago
  16. FoolForMath Group Title
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    Yes, but it depends on the problem again we can call it whole too.

    • 2 years ago
  17. waheguru Group Title
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    Do i get a metal for asking a good question =)

    • 2 years ago
  18. Jemurray3 Group Title
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    We could also call it prime, which is even more precise. I disagree that it's always better to specify the subset rather than the superset. In that case, the best classification of 2 would be "the number 2". Regardless. The question was answered, the points were made. End of thread. And sure^, why not.

    • 2 years ago
  19. waheguru Group Title
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    yaay im level 18

    • 2 years ago
  20. waheguru Group Title
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    no 20

    • 2 years ago
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