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Proof #2

  1. Apr 29, 2007 #1
    If a circle has an area of t square units and radius r, then a circle with area 2t square units has a radius 2r

    a) State whether or not the statement is true, Justify your answer.

    b) If it is not true, rewrite the statement so that it is true.


    50 = (4)(4)(pi)
    100 = (5.6)(5.6)(pi)

    5.6/2 does not equal 4, therefore the statement is false.

    If a circle has an area of t square units and radius r, then a circle with area 4t square units has a radius of 2r.

    50 = (4)(4)(pi)
    200 = (8)(8)(pi)

    Does this look ok?
    I know, I ask a lot of questions but I want to make sure I understand this stuff
     
  2. jcsd
  3. Apr 29, 2007 #2

    Dick

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    Aside from the fact that at some point you should have enough confidence to stop asking questions and start trusting yourself. It looks GREAT.
     
  4. Apr 29, 2007 #3

    Dick

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    Well, and 4*4*pi does not equal 50. I think you are more right in spirit than in literal fact.
     
    Last edited: Apr 29, 2007
  5. Apr 29, 2007 #4
    4pi^2 = 50 to two sig. fig.

    That's good enough for government work.
     
  6. Apr 29, 2007 #5

    Dick

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    True. Just experimenting with the numbers is a great first step. But you did mean 4^2*pi, right?
     
  7. Apr 29, 2007 #6
    Yeah, I rounded off the numbers so the example wouldn't be overly messy and since the proof is what the question is really asking for.

    Thanks again guys
     
  8. Apr 30, 2007 #7

    Integral

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    You really do not need to, and should not, use specific numbers in this type of problem.
    you are given this to be true:
    [tex] t = \pi r^2 [/tex]

    now what is the area for a circle with radius 2r?

    [tex] \pi (2r)^2 = 4 \pi r^2 = 4t \neq 2t [/tex]

    So the given statement is false.
     
  9. Apr 30, 2007 #8

    Hurkyl

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    This is in a geometry class, right? I strongly suspect that whoever wrote the question did not intend for you to invoke the area formula of a circle. (Unless they're building up to the easy way to do this problem)
     
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