waheguru
  • waheguru
by connecting the midpoints of a reatangle u get a rhombus right?
Mathematics
  • Stacey Warren - Expert brainly.com
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
yes.
anonymous
  • anonymous
yup
anonymous
  • anonymous
In general you get a parallelogram, which does not necessarily have to be a rhombus.

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anonymous
  • anonymous
No you will get a rhombus always.
anonymous
  • anonymous
and square gives square.
anonymous
  • anonymous
Oh my, I was thinking of something else, I apologize. Yep :)
anonymous
  • anonymous
A square is a rhombus though.
anonymous
  • anonymous
No problem :)
anonymous
  • anonymous
If you join the midpoints of a square the secondary figure is also a square.
anonymous
  • anonymous
Yes, but I meant that a square is also a rhombus.
anonymous
  • anonymous
Every square is rhombus, rectangle and parallelogram.
anonymous
  • anonymous
I disagree. A rhombus is a quadrilateral with all four sides being of the same length. A square is an example of this.
anonymous
  • anonymous
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.
anonymous
  • anonymous
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.
anonymous
  • anonymous
Why not natural then?
anonymous
  • anonymous
Yes, but it depends on the problem again we can call it whole too.
waheguru
  • waheguru
Do i get a metal for asking a good question =)
anonymous
  • anonymous
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.
waheguru
  • waheguru
yaay im level 18
waheguru
  • waheguru
no 20

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