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-   -   Broken plate math problem (https://www.askmehelpdesk.com/showthread.php?t=830464)

  • Feb 8, 2017, 11:05 AM
    aaknutsen
    Broken plate math problem
    I was wondering if it was possible to help me with finding different difficult methods to finding out the circumference and diameter of a broken plate if you only have half. I have used a method using perpendicular bisectors to find the middle of the plate and than times by 2 two get the full diameter. I also plan on using pythagoras theorem.. Are there any other methods I could use? :)
  • Feb 8, 2017, 12:05 PM
    ma0641
    With a round plate, If you have 1/2 of the plate, you have "r". What would ," d" be? Formula for circumference is? X "d". Unless your plate is square, Pythagorean theorem won't work.
  • Feb 9, 2017, 03:31 AM
    aaknutsen
    Hi, thank you for your help on that.. I need possibly further methods I can use though, exactly for that purpose.
  • Feb 9, 2017, 04:46 AM
    joypulv
    I assume that you don't really mean 'half' a plate? That would be too easy.

    https://www.geogebra.org/m/m97sTNZH (click on solution boxes)

    I have a feeling you have done this. I don't know of any other ways.
  • Feb 9, 2017, 11:06 AM
    aaknutsen
    I do mean half of a plate..

    It is this problem here: http://mathforum.org/pow/teacher/samples/MathForumSampleGeometryPacket.pdf

    I mean I need help in more ways than what is desrcibed in the website above. I need methods such as creating equations and so on if that is possible...
  • Feb 9, 2017, 01:39 PM
    joypulv
    That does not deal with half a plate, but a broken piece of unknown fraction of the whole. The typical 'broken plate' problem.

    http://mathforum.org/pow/teacher/sam...etryPacket.pdf

    I posted the link again because most responders here won't open links. It's OK.

    My link was a good one. What don't you like about it?
  • Feb 9, 2017, 02:26 PM
    aaknutsen
    Thank you for replying! Well yes, I don't mean a perfect half of a plate obviously, that would be too easy.. I am very unfamiliar with geogebra as I am unsure how it can help me with finding different methods I can use.. This is on me of course. I thought that the program was just used for creating a visual representation of math problems.
  • Feb 13, 2017, 11:25 AM
    ebaines
    I think you are using the best approach, if I understand you correctly. As the site that Joy referenced explained, you draw two chords across any portions of the edge of the plate, then draw perpendicular bisectors of those chords. The point at which those two bisectors intersect is the center of the circular plate. Then measure the distance to the edge to get the radius.

    You can do all this mathematically with equations. The process would be:

    1. Select points A, B, and C. My advice would be to establish a coordinate system that makes these points "easy" to deal with, but it doesn't really matter.

    2. Determine the slope for Chord 1 (connecting points A and B) and Chord 2 (connecting points B and C).

    3. Find the midpoint of both chords. The x-coordinates of the midpoint of Chord 1 is the midpoint between the x-coordinates of points A and B and the y-coordinate of the midpoint of Chord 1 is the midpoint between the y-coordinates of points A and B. Use the same process to find the midpoint coordinate of Chord 2.

    4. Determine thee equations of the bisectors of Chord 1 and the bisector of Chord 2. The bisector runs through the midpoints that you just determined, and the slope of each bisector is the negative reciprocal of the chord it bisects (that's' why we needed to find the slope if the chords in step 2). Using the form y = mx+b you have values for x, y, and m, so you can solve for b.

    5. Determine the point at which the two bisectors intersect.

    6. Calculate the distance from that point to either A, B, or C using the Pythagorean theorem. It doesn't matter which point A, B or C you pick - all three should yield the same distance from the center. But to check your work I suggest doing this calculation with at least two of these point, if not all three.

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