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by Gb_A on October 6th, 2011

Gb_A

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A rectangle ABCD of dimension r and 2r is folded along the diagonal BD such that ABD and CBD are perpendicular and distance AC' is?

where C' is the new position of C. Answer in terms of 1/r..

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Answers. 2 helpful answers below.

  • by WarHorseLeBron on October 6th, 2011

    WarHorseLeBron

    Dude, what does it mean to say that two angles are perpendicular? When you fold a rectangle on its diagonal, there's nothing perpendicular about it (unless its a square)...

    ...So, the source of my inspiration for this answer is a piece of notebook paper that I folded on its diagonal. It looks like two triangular mountains, one in the foreground and one behind it, with the tops of the moutains being points A and C

    The diagonal length BD = sqrt(5r²) = r*sqrt(5)

    The two halves of the triangles both have a height, h.

    A = 2r²/2 and A = bh/2 so h = 2r²/b = 2r/sqrt(5).

    Now, how far along the diagonal is this from point B or point D?

    r²-h² = x² = r²-4r²/5 = 0.2r²

    x = r*sqrt(0.2)

    Multiply x by 2 and subtract from the total length BD

    r*sqrt(5)-2r*sqrt(0.2) = r*sqrt(1)-2r*(0.2) = r-0.4r = 0.6r

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  • by iwnit on October 6th, 2011
    voted: None of these.

    iwnit

    1) Let's first consider the original rectangle, and let's call:
    - A1 the projection of A on BD,
    - C1 the projection of C on BD
    (I strongly suggest to draw a picture to follow the demonstration)

    According to Pythagoras:
    BD^2 = AB^2 + AD^2 = (2^2 +1^2)*r^2 = 5*r^2
    The negative value is not possible, so:
    BD = sqrt(5) * r

    There are two way of calculating the suface od the triangle ABD:
    AB*AD/2 = BD*AA1/2
    So:
    AA1 = AB*AD/BD = (r*2*r)/(sqrt(5) * r) = [2/sqrt(5)]*r

    Let's also notice that:
    AA1 = C1C = C1C'

    According to Pythagoras:
    C1D^2 = CD^2 - C1C^2 = r^2 - (4/5)*r^2 = (1/5)*r^2
    The negative value is not possible, so:
    C1D = [1/sqrt(5)]*r
    (this is also the value of A1B)

    Let's calculate A1C1:
    A1C1 = BD - 2*C1D = sqrt(5) * r - [2/sqrt(5)]*r = {sqrt(5) - [2/sqrt(5)]}*r = [(5 - 2)/sqrt(5)]*r = [3/sqrt(5)]*r

    Now we can apply Pythagoras to the rectangles AC1C' and AA1C1:
    AC'^2 = AC1^2 + C1C'^2
    AC1^2 = AA1^2 + A1C1^2
    And combining both:
    AC'^2 = AA1^2 + A1C1^2 + C1C'^2

    We know that C1C' = AA1, so we can write:
    AC'^2 = 2*AA1^2 + A1C1^2

    We have calculated AA1 and A1C1 above:
    AC'^2 = 2*(4/5)*r^2 + (9/5)*r^2 = (17/5)*r^2
    The negative value is not possible, so:
    AC' = sqrt(17/5)*r = [sqrt(17*5)/5]*r = [85^(1/2)/5]*r

    This result is somewhat similar to the fourth choice. However, the writing in the fourth choice would normally imply that r is under the fraction bar, like in the first three choices.

    Please notice that this solution would be right if you were using a BIG fraction bar, like
    85^(1/2)
    -------------- * r
    5


    2) From the point of view of dimensional analysis, it does not make sense for r to be under the fraction bar in the solution. So neither any of the first four choices does makes sense, nor the proposition "Answer in terms of 1/r".
    Further information:
    http://en.wikipedia.org/wiki/Dimensional_analysis

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