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Math Help - Finding the value of "x" inside a square

  1. #1
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    Finding the value of "x" inside a square

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    Last edited by Macleef; November 28th 2007 at 08:57 PM.
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  2. #2
    Senior Member DivideBy0's Avatar
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    This can be solved using simultaneous equations. Please refer to the diagram for the labelling.

    From the diagram we can see (by Pythagoras' Theorem):

    a^2+b^2 = 7^2 ...[1]

    b^2 + c^2 = 35^2 ...[2]

    c^2+d^2 = 49^2 ...[3]

    a^2+d^2 = x^2 ...[4]

    Subtracting the first from the second,

    c^2 - a^2 = 35^2 - 7^2

    c^2 - a^2 = 1176 ...[5]

    Then, subtracting [5] from [3],

    a^2 + d^2 = 49^2 - 1176

    So we have x^2 = 49^2 - 1176=1225

    x = 35
    Attached Thumbnails Attached Thumbnails Finding the value of "x" inside a square-distances.jpg  
    Last edited by DivideBy0; November 28th 2007 at 04:53 PM.
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  3. #3
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    Wow...I'm so confused -_-
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  4. #4
    Senior Member DivideBy0's Avatar
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    Ok, could you please tell me which part confused you? Do you understand the diagram, have you learnt Pythagoras' Theorem?
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  5. #5
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    Quote Originally Posted by DivideBy0 View Post
    Ok, could you please tell me which part confused you? Do you understand the diagram, have you learnt Pythagoras' Theorem?
    I don't get why a^2 + b^2 = 7^2. If this is the Pythagorean theorem, how did you come to that equation? I mean, when I physically look at the diagram, those 2 sides, "a" and "b" are not in the same triangle... I don't know if I'm clearly explaining myself here...





    What happens to the other 2 sides that aren't labelled?
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  6. #6
    Senior Member DivideBy0's Avatar
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    You can make them be in the same triangle :P, since what you have here is a rectangle, the opposite sides are equal.
    Attached Thumbnails Attached Thumbnails Finding the value of "x" inside a square-6sh1ilc.jpg  
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  7. #7
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    Thanks, that clears up everything, I understand it now

    And I think what I did previously is kind of similar to the process you posted
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