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Math Help - Ratio of area of two circles

  1. #1
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    Exclamation Ratio of area of two circles

    A circle with center O intersects another circle with center P at points C and D. if angle COD = 30 and angle CPD = 60, what is the ratio of the area of the circle with center O to the area of the circle with center P?
    Answer= 1: (2-√3)
    Can someone paint me a picture of this because I don't completely understand this question.
    Last edited by mr fantastic; December 12th 2009 at 02:00 AM. Reason: Changed post title
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  2. #2
    Super Member 11rdc11's Avatar
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    Quote Originally Posted by shane99 View Post
    A circle with center O intersects another circle with center P at points C and D. if angle COD = 30 and angle CPD = 60, what is the ratio of the area of the circle with center O to the area of the circle with center P?
    Answer= 1: (2-√3)
    Can someone paint me a picture of this because I don't completely understand this question.
    Are you sure it is not?

    2+\sqrt{3}
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    Yes, I am 100% sure. (lol the answer is actually in the back of the book)
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    Super Member 11rdc11's Avatar
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    Quote Originally Posted by shane99 View Post
    Yes, I am 100% sure. (lol the answer is actually in the back of the book)
    O then I'm not sure but I can tell you how I approached the problem. First draw out the problem.

    Then I gave line CD value of 1.

    now using this info and that angle cpd is 60 degrees I figure out the radius for the circle with center is 1. Then I used the equation to find area of a circle and got the area to be 1pi.

    Now I use the same idea for the other circle. To find the radius of the other circle is a bit harder so I used the law of sines

    \frac{\sin{30}}{1} = \frac{\sin{75}}{r}

    Can you finish up from here
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  5. #5
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    Quote Originally Posted by shane99 View Post
    A circle with center O intersects another circle with center P at points C and D. if angle COD = 30 and angle CPD = 60, what is the ratio of the area of the circle with center O to the area of the circle with center P?
    Answer= 1: (2-√3)
    Can someone paint me a picture of this because I don't completely understand this question.
    connect chord CD

    OC = OD = R

    PC = PD =  CD = r

    cosine law ...

    r^2 = R^2 + R^2 - 2R^2\cos(30)

    r^2 = 2R^2[ 1 - \cos(30)]

    r^2 = 2R^2\left(\frac{2-\sqrt{3}}{2}\right)

    r^2 = R^2(2-\sqrt{3})

    \frac{1}{2-\sqrt{3}} = \frac{R^2}{r^2}

    \frac{1}{2-\sqrt{3}} = \frac{\pi R^2}{\pi r^2}
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  6. #6
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    Quote Originally Posted by 11rdc11 View Post
    Are you sure it is not?

    2+\sqrt{3}
    11rdc11:

     2+\sqrt{3} : 1 = 1: 2- \sqrt{3}

    Your method is also correct, just different ways of expressing the ratio.
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  7. #7
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    Hello, shane99!

    A circle with center O intersects another circle with center P at points C and D.
    If \angle COD = 30^o and \angle CPD = 60^o,
    what is the ratio of the area of the circle O to the area of the circle P?

    Answer: . 1: (2-\sqrt{3})
    Here's the diagram ... without the circles.

    Code:
                          C
                          *
                 R    *   | *
                  *       |   * r
              *           |     *
        O *  30         d|   60 * P
              *           |     *
                  *       |   * r
                 R    *   | *
                          *
                          D

    The radius of circle O is R.
    The radius of circle P is r.

    The ratio of the areas is: . \text{Ratio} \:=\:\frac{\pi R^2}{\pi r^2} \;=\;\frac{R^2}{r^2}


    Apply the Law of Cosines to \Delta OCD and \Delta PCD:

    . . R^2 + R^2 - 2R^2\cos30^o \:=\:d^2 \:=\:r^2 + r^2 - 2r^2\cos60^o

    . . 2R^2(1-\cos30^o) \;=\;2r^2(1-\cos60^o) \quad\Rightarrow\quad \frac{R^2}{r^2} \;=\;\frac{1-\cos60^o}{1-\cos30^o}

    Therefore: . \text{Ratio} \;=\;\frac{R^2}{r^2} \;=\;\frac{1 - \frac{1}{2}}{1-\frac{\sqrt{3}}{2}} \;=\;\frac{2}{2}\cdot\frac{1-\frac{1}{2}}{1-\frac{\sqrt{3}}{2}} \;=\; \frac{1}{2-\sqrt{3}}


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