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Math Help - A geometry tutorial

  1. #1
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    An example from a math book

    I saw an example from math book that I don't really realize. If you look at attachment, then you see a drawing related to this example. There are two circle which are in the same size. Both radius are 12 cm. The points O and O1 are centers of the circles. If you look at attachment then you see that there are formed two equilateral triangle OO1B and OO1A. My task is to find a general area (the area which is between the points BOA and BO1A)

    My math book says that better is at first to find half of the area. We get this when we disjoint an area of a sector OAB by(from?) an area of a triangular AOB. Because sector's angle is 120 degrees, then sector's area is: 1/3*pi*12^2=48*pi(cm^2)

    Area(OAB)=[12^2*sqrt(3)]/4=36sqrt(3)(cm^2)

    And according to this the required area = 2* [48*pi -36sqrt(3)]

    Answer: The general area (the area which is between the points AOBO1) is 2* [48*pi -36sqrt(3)]

    The following things need an explanation:
    1) How to show that sector's angle is 120 degrees.
    2) Don't understand that one: [48*pi -36sqrt(3)]

    This approach seems to me confusing.
    Please explain me. Thanks in advance.
    Attached Thumbnails Attached Thumbnails A geometry tutorial-kaks_ringi.jpg  
    Last edited by totalnewbie; August 6th 2005 at 10:33 AM.
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  2. #2
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    The picture shows that the distance OO1 is 6+6=12 and the radius of each circle = OA = O1A = 12. So the triangle OO1A is equilateral and the angle O1OA is 60 degrees. By symmetry so is the angle O1OB and so AOB is 120 degrees.

    Now let's consider the area of the intersection on the right of the line AB.
    (You'll want twice this in the end.) This is a sector of a circle with a triangle removed: specifically, the sector of the circle centre O between the radii OA and OB, minus the triangle ABO.

    The area of a sector of a circle of radius 12 and angle 120 degrees is one-third (120/360) of the area of the whole circle which is pi.12^2: so pi.12^2/3 = 48.pi

    The area of a triangle with angle 120 and sides 12, 12 enclosing that angle is computed as follows. It is isosceles and has the other angles 30,30. So the height (from O down to AB) is 12 sin 30 = 12. 1/2 = 6. The base is twice 12 cos 30 = 2.12.sqrt(3)/2 = 12 sqrt(3). The area is half base times height = (1/2).6.12.sqrt(3) = 36 sqrt(3).

    Putting these together gives you the area you already mentioned.
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  3. #3
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    A sector is an area cut out by two radii from the centre: its area is d/360 times the area of the circle where d is the angle in degrees between the two radii. So in your coloured diagram, the green and pink areas together constitute a sector of the circle on the right.
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  4. #4
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    First of all I deleted my previous post. I am not able to imagine how I could reckon something like on my deleted post. Sorry.

    I took out my dividers and a piece of paper and I realize now that I have misunderstood all that stuff. It's all because of misunderstanding the sector concept.
    Meanwhile I was thinking that I'm stupid because I don't realize so simple thing. Now I really have to admit that all the concepts are important in math.
    Last edited by totalnewbie; August 7th 2005 at 02:23 AM.
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