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  1. #1
    Web Ho - Design B!tch ~Michelle's Avatar
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    Red face Geometry Help Please - Anyone good at it?
    Hoping someone here can help.

    Trying to help my daughter with a Geometry problem (which I totally sucked at in school, by the way) and I can't wrap my head around this problem at all.

    Can somone give it a shot and explain to me how to do this?

    Here is the question/problem:

    A radius of a circle, OC is perpendicular to chord AB and intersects it at point M. Given that AM = 45-5x and BM = 30+10x, solve for x and find the lengths at AM and BM.

    TIA!
    ~Michelle
    "All I ask is a chance to prove that money can't make me happy."
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  2. #2
    Life is Supposed to be Fun! Rexanne's Avatar
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    OMG WFT?!! No way - my kids are on their own with this kind of insane crap. LMAO

    I figured out that x=40 - after that, my brain couldn't possibly wrap around it either. Good luck. Someone should be along soon with a math brain and the answer.
    Peace,

    Rexanne

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    Loving Everyone's Child Creates Magic


  3. #3
    Affiliate Manager Matt McWilliams's Avatar
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    Quote Originally Posted by ~Michelle
    Hoping someone here can help.

    Trying to help my daughter with a Geometry problem (which I totally sucked at in school, by the way) and I can't wrap my head around this problem at all.

    Can somone give it a shot and explain to me how to do this?

    Here is the question/problem:

    A radius of a circle, OC is perpendicular to chord AB and intersects it at point M. Given that AM = 45-5x and BM = 30+10x, solve for x and find the lengths at AM and BM.

    TIA!
    OK, it has been a LOOONNNG time since I took geometry but...

    If the lines are perpendicular, then AM = BM

    Thus, 45-5x = 30 + 10x

    Thus x = 1

    AM and BM both = 40

    Unless I am wrong, then I am right
    Matt McWilliams
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  4. #4
    Life is Supposed to be Fun! Rexanne's Avatar
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    Wow - I'm impressed Matt!
    Peace,

    Rexanne

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    Loving Everyone's Child Creates Magic


  5. #5
    Web Ho - Design B!tch ~Michelle's Avatar
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    *LOL* at you guys!

    Thanks!

    I think Matt might be right. That is what my husband just came up with but he wasn't positive. So, if two of you (he and you) came up with the same thing, than that is good enough for me! *LOL*

    Warning: There are a couple more real tough ones we are working on right now, and if we don't get them, I will be back!
    ~Michelle
    "All I ask is a chance to prove that money can't make me happy."
    "Work to become, not to acquire." -- Confucius

  6. #6
    Affiliate Manager Matt McWilliams's Avatar
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    10 years later and I FINALLY make my geometry professor proud.

    Here's to Mrs. Tyree...if only she could see me now
    Matt McWilliams
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  7. #7
    Moderator MichaelColey's Avatar
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    x=3
    Michael Coley
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  8. #8
    Web Ho - Design B!tch ~Michelle's Avatar
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    I have to say Matt, you da man!

    I HATE, and I mean HATE math in general and when it comes to Geometry, Algebra and Trig, I get a huge headache just thinking about doing it!

    My daughter took the 1st half of Geometry this year in class with a real live teacher, but in order to fill one of her other 1/2 credits in another subject she needed, she had to switch to an online class for her second half of Geometry for the last half of the year. Well, the online class was further ahead of what she was doing in the first half of the year, so we have spent the past 3 days, helping her get caught up to where she at least understands what she is supposed to be doing.

    She has about another 1/2 hr or so to go tonight, then we are quiting for today. A few more assignments tomorrow and she should be good to go come Monday.

    I have to tell ya, I would rather poke myself in the eye with a rusty nail then do Geometry!
    ~Michelle
    "All I ask is a chance to prove that money can't make me happy."
    "Work to become, not to acquire." -- Confucius

  9. #9
    Web Ho - Design B!tch ~Michelle's Avatar
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    Quote Originally Posted by MichaelColey
    x=3
    Please tell me you are joking! *LOL*

    If not, how did you arrive at that?
    ~Michelle
    "All I ask is a chance to prove that money can't make me happy."
    "Work to become, not to acquire." -- Confucius

  10. #10
    Affiliate Manager Matt McWilliams's Avatar
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    I am generally pretty good at math...not so much with Calculus though.

    I have always had an "eye" for numbers...one of my professors said that. I can see things and get them quickly for some reason.

    So good at math...but I am probably the slowest reader alive today.

    If you need anymore help, you know where to reach me
    Matt McWilliams
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  11. #11
    Moderator MichaelColey's Avatar
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    Oops, I wrote it down and solved it and couldn't read my own writing. I mistook a + for a -.

    x does equal 1.

    45-5x = 30+10x
    15-5x = 10x (subtract 30 from each side)
    15 = 15x (add 5x to each side)
    x = 1 (divide by 15)
    Michael Coley
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  12. #12
    Web Ho - Design B!tch ~Michelle's Avatar
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    Whew! I almost went into a panic because I "thought" I understood what Matt said. *LOL*



    Quote Originally Posted by MichaelColey
    Oops, I wrote it down and solved it and couldn't read my own writing. I mistook a + for a -.

    x does equal 1.

    45-5x = 30+10x
    15-5x = 10x (subtract 30 from each side)
    15 = 15x (add 5x to each side)
    x = 1 (divide by 15)
    ~Michelle
    "All I ask is a chance to prove that money can't make me happy."
    "Work to become, not to acquire." -- Confucius

  13. #13
    Pimp Duck popdawg's Avatar
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    See spot run
    ================================================================
    Been away, now I'm back. Not as much, but I'm back & starting from scratch. Where I was, was fantastic. Where I am now, less so. Things have changed, become harder. So have I. Game ON!!!
    ================================================================

  14. #14
    Full Member ske9963's Avatar
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    Quote Originally Posted by MattMcWilliams
    If the lines are perpendicular, then AM = BM
    I do not think that is true.

    perpendicular = 90' from the line.

    So AM = BM can be true but not necessary.
    Ma, where the beer? :escape:

  15. #15
    Moderator MichaelColey's Avatar
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    The radius comes straight out from the center. A chord that is perpendicular to that radius would have to be equidistant from the radius to the two points on the circle, otherwise it wouldn't be perpendicular.
    Michael Coley
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  16. #16
    Affiliate Network Rep Kim Salvino's Avatar
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    This is just my personal opinion, but I think geometry should count as two credits - math and foreign language!
    Kim Salvino, Client Services Director, Performance Horizon Group
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  17. #17
    Life is Supposed to be Fun! Rexanne's Avatar
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    Geeeeeeeze you guys ... my head is starting to hurt trying to decipher what everyone's saying. Yeah, Kim ... a foreign language, indeed!
    Peace,

    Rexanne

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    Loving Everyone's Child Creates Magic


  18. #18
    Newbie laguna3dguy's Avatar
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    Here's a Visual Aid
    I needed to visualize it and try to remember exactly what a chord was.
    .

    The key to the problem is knowing that AB is perpendicular to the radius.
    The radius, by definition, originates at the center, therefore, no matter where point M ends up along OC, AM must equal BM.

    It's true. Math is a language. Knowing the terms is half the battle.
    Last edited by laguna3dguy; March 4th, 2007 at 05:28 AM. Reason: addendum and text styling

  19. #19
    Troll Killer and best Snooper!
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    Isn't it amazing how having a picture of the problem can make all the difference?

    I was totally lost until that picture was posted.

    Geometry without pictures is evil.

  20. #20
    Web Ho - Design B!tch ~Michelle's Avatar
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    Stuck on another Geometry Problem
    We are stuck on one more problem. After this one she will be all caught up.

    Question is:

    Solve for x, given that EF is the median of trapezoid ABCD. Show your work.



    Anyone want to give it a shot?
    ~Michelle
    "All I ask is a chance to prove that money can't make me happy."
    "Work to become, not to acquire." -- Confucius

  21. #21
    Affiliate Manager Matt McWilliams's Avatar
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    Quote Originally Posted by ske9963
    I do not think that is true.

    perpendicular = 90' from the line.

    So AM = BM can be true but not necessary.
    Here is a visual that may help. No matter where AB is, AM and BM will always be the same
    Attached Images Attached Images
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  22. #22
    Advocate mellie's Avatar
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    For the trapezoid problem-

    I only have couple minutes but this should help. The length of EF is the average of the other two bases so ( 12 + (3x-3) )/2 = 2x + 1 or length AD + length BC divided by 2 equals length of EF. In a trapezoid the median has a length that is equal to the mean of the lengths of it's bases.
    Last edited by mellie; March 4th, 2007 at 02:26 PM. Reason: Added that this is for trapezoid problem.
    Melanie
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  23. #23
    Affiliate Manager Matt McWilliams's Avatar
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    Quote Originally Posted by ~Michelle
    We are stuck on one more problem. After this one she will be all caught up.

    Question is:

    Solve for x, given that EF is the median of trapezoid ABCD. Show your work.



    Anyone want to give it a shot?
    OK, the median of a trapezoid = the mean of the lengths of the two bases...

    So

    (12 + (3x - 3) / 2) = 2x + 1

    Thus x = 7

    12 + (21-3) / 2 = 15
    14 + 1 = 15

    I forget how to do the math on it. The numbers just jump out at me. It used to piss off my math teachers.
    Matt McWilliams
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  24. #24
    Life is Supposed to be Fun! Rexanne's Avatar
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    Quote Originally Posted by MattMcWilliams
    No matter where AB is, AM and BM will always be the same
    That just sounds WRONG!
    Peace,

    Rexanne

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    Loving Everyone's Child Creates Magic


  25. #25
    Affiliate Manager Matt McWilliams's Avatar
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    Rex, you just have such a DIRTY mind
    Matt McWilliams
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