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Using the Intermediate Value Theorem

  1. Oct 2, 2009 #1
    Show that there is a square with a diagonal that is between r and 2r and an area that is half the area of a circle of radius r.
    The Theorem: If f is continuous on a closed interval [a,b] and k is any number between f(a) and f(b), inclusive, then there is at least one number x in the interval [a,b] such that f(x)=k.
    Let l be a length of a side of the square, then the diagonal's length is [tex] \sqrt{2}l \mbox{ and } r < \sqrt{2}l < 2r \mbox{ that implies } f(a)=f(r) = r^2 \mbox{ and } f(b)=f(2r)= 4r^2 [/tex]. Can I find [tex] k = \frac{\pi}{2}r^2 [/tex]? I just cannot visualize k on the graph of [tex] l^2[/tex] and in between f(a) and f(b), because it is impossible. So where am I going wrong? I'm doing this for my own interest. Please help. Thank you.
     
  2. jcsd
  3. Oct 2, 2009 #2

    LCKurtz

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    The area of the square with diagonal r is r2/2, and the area of the square with diagonal 2r is 2r2. The area of the half circle is πr2/2, which is between the areas of the squares.
     
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