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Math Help - A Rectangular Piece of Paper is folded

  1. #1
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    A Rectangular Piece of Paper is folded

    The Problem: A rectangular piece of paper, ABCD, is folded so that point B lies on point D. It is then unfolded to show the crease XY. If AX = 6 and XB = 10, find the length of the crease XY.


    10 X 6
    B________________________ A
    | / |
    | / |
    | / |
    | ________/______________ |
    C Y D

    I realize that both AB and CD are 16 across. I've tried turning parts of this rectangel into a triangle. But that didn't yield results, as I couldn't determine the new figure's length. The closest I got to a lead was My thought of drawing a line from Y to A. That would give me a triangle with a base of 6, but no angles, so no way to determine XY.
    Last edited by D. Martin; June 1st 2008 at 02:11 PM. Reason: Removed the "Note", which was a mistake to put in. I tried to fix up the diagram.
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  2. #2
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    I just added the diagram. I couldn't make it on this form.
    Attached Thumbnails Attached Thumbnails A Rectangular Piece of Paper is folded-figure-problem.bmp  
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  3. #3
    Member Jonboy's Avatar
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    Since you have a rectangle, the length is half the width. So BC and AD are 8.

    And you can make this picture: link

    Now apply the Pythagorean Theorem a bunch of times.

    (XB)^2 + (BC)^2 = (XC)^2

    8^2 + 10^2 = (XC)^2

    164 = (XC)^2

    XC = 2\sqrt{41}

    Then apply it again: (XA)^2 + (AD)^2 = (DX)^2

    6^2 + 8^2 = (DX)^2

    36 + 64 = (DX)^2

    80 = (DX)^2

    DX = 4\sqrt{5}

    Now let's fill that in our pic: link

    Since XZ is an angle bisector, then ZC = DZ

    So both are 8.

    Apply Pythagorean Theorem to get XZ

    (ZC)^2 + (XZ)^2 = (2\sqrt{41})^2

    64 + (XZ)^2 = 164

    (XY)^2 = 100

    XY = 10

    And without more info, I don't see how you can find XY
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    Thank you

    Thank you, very much. That was really helpful and I wouldn't have figured out the problem by myself before tomorrow. So I really appreciate your assistance.
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  5. #5
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    Quote Originally Posted by Jonboy View Post
    Since you have a rectangle, the length is half the width. So BC and AD are 8.

    And you can make this picture: link

    Now apply the Pythagorean Theorem a bunch of times.

    (XB)^2 + (BC)^2 = (XC)^2

    8^2 + 10^2 = (XC)^2

    164 = (XC)^2

    XC = 2\sqrt{41}

    Then apply it again: (XA)^2 + (AD)^2 = (DX)^2

    6^2 + 8^2 = (DX)^2

    36 + 64 = (DX)^2

    80 = (DX)^2

    DX = 4\sqrt{5}

    Now let's fill that in our pic: link

    Since XZ is an angle bisector, then ZC = DZ

    So both are 8.

    Apply Pythagorean Theorem to get XZ

    (ZC)^2 + (XZ)^2 = (2\sqrt{41})^2

    64 + (XZ)^2 = 164

    (XY)^2 = 100

    XY = 10

    And without more info, I don't see how you can find XY
    I'm just wondering, how did you deduce that the length was half the width?
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  6. #6
    Member Jonboy's Avatar
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    Honestly, I assumed too much.

    The only way I could've got the width was knowing the perimeter, which we don't know.

    That's because:

    P = 2L + 2W

     P = 2(16) + 2W

     2W = P - 32

     W = \frac{P - 32}{2} = \frac{1}{2}P - 16

    But, look on the bright side, it further proves you can't solve this problem.
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  7. #7
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by D. Martin View Post
    The Problem: A rectangular piece of paper, ABCD, is folded so that point B lies on point D. It is then unfolded to show the crease XY. If AX = 6 and XB = 10, find the length of the crease XY.


    10 X 6
    B________________________ A
    | / |
    | / |
    | / |
    | ________/______________ |
    C Y D

    I realize that both AB and CD are 16 across. I've tried turning parts of this rectangel into a triangle. But that didn't yield results, as I couldn't determine the new figure's length. The closest I got to a lead was My thought of drawing a line from Y to A. That would give me a triangle with a base of 6, but no angles, so no way to determine XY.
    I don't have time to get to it right now, but I'll try (no promises) to get back to it later.

    Note that whereas we don't know the width of the rectangle, we do know that points B and D lie the same distance away from the fold along the red line (the diagonal) because when the paper is folded these points overlap. My diagram stinks, but it gives you the idea behind what I'm talking about anyway.

    Edit: Oh! I almost forgot. BD and XY have to be perpendicular as well. Do you see why that is?

    -Dan
    Attached Thumbnails Attached Thumbnails A Rectangular Piece of Paper is folded-box.jpg  
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  8. #8
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    Quote Originally Posted by D. Martin View Post
    The Problem: A rectangular piece of paper, ABCD, is folded so that point B lies on point D. It is then unfolded to show the crease XY. If AX = 6 and XB = 10, find the length of the crease XY.


    10 X 6
    B________________________ A
    | / |
    | / |
    | / |
    | ________/______________ |
    C Y D
    I've sketched the rectangle.

    BX = XD = 10 and XA = 6. Use Pythagorean theorem in the triangle XAD. Therefore AD = 8.

    Use Pythagorean theorem with the grey triangle to calculate the length of the crease. I've got c = \sqrt{16+64} = 4 \cdot \sqrt{5}
    Attached Thumbnails Attached Thumbnails A Rectangular Piece of Paper is folded-papierfalten.gif  
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  9. #9
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    I'd forgotten how much I love this stuff.
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