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Math Help - 4 urgent circumference and areas

  1. #1
    Junior Member
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    Question 4 urgent circumference and areas

    My son is almost done with this course, I just need a little more help explaining this stuff to him please!!!!!

    I think the last one is 99pi.
    For the question with the trapezoid, do I take 32-14=18 to try to get the areas of the shaded parts? and if so, how do i decide how long each side is?
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    Last edited by sanee66; July 20th 2007 at 05:15 PM. Reason: checking an answer
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  2. #2
    Senior Member DivideBy0's Avatar
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    Question 2

    A_{shaded}=\frac{1}{2} \times 11 \times (32-14)
    A_{shaded}=99

    Question 3

    HD=DF=GF=HG=32\pi

    Then, by pythagoras' formula:

    \left(\frac{HG}{2} \right)^2 + \left(\frac{GF}{2} \right)^2=IJ^2

    \sqrt{(16\pi)^2+(16\pi)^2}=IJ

    \sqrt{512\pi^2}=IJ


    IJ=16\sqrt{2}\pi

    Question 4

    2\pi r_{A}=24\pi
    r_{A}=12
    \pi r_{A}^2=144\pi


    2\pi r_{B}=12\pi
    r_{B}=6
    \pi r_{B}^2=36\pi

    Difference is 144\pi-36\pi=108\pi

    Sorry I can't do Q.1, are you sure there is enough information?
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