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Math Help - Parallellogram - Euclidean Geometry

  1. #1
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    Parallellogram - Euclidean Geometry

    Okay I'll just ask the question.

    If I know that the sides of a parallellogram is 1 (l.u = length unit ) and 2 l.u and that one angle is 60degrees, can I using plain Euklides Geometry calculate the diagonals exactly?

    No need for you guys to solve it completly but a tip would be nice!

    I can get all angles easy. This would be a breeze if it was a rectangle, but cuz it's leaning the diagonal must become somewhat longer. I tried to get it by creating it into a 90degree traingle so I could use phytagoras, but for that I'd have to make a million small triangles and I hardly belive it's the easiest way.

    Thanks in advance.
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  2. #2
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    Hello, Hanga!

    If I know that the sides of a parallellogram is 1 and 2, and that one angle is 60,
    can I, using plain Euclidean Geometry, calculate the diagonals exactly? . . . . Yes!

    We have parallelogram ABCD with: . AB = CD = 2,\;AD = BC = 1,\;\angle B = \angle D = 60^o

    This parallelogram can be divided into four equilateral triangles!


    Code:
              A * - - - - - * - - - - - * B
               / \         / \         /:
              /   \       /   \       / :   _
           1 /     \     /     \     /  : √3
            /       \   /       \   /   :
           / 60     \ /         \ /    :
          * - - - - - * - - - - - * - - +
          D     1           1     C    E

    Draw diagonal AC.

    It can be seen that \Delta DAC is a right triangle
    . . with: \angle DAC = 90^o,\:AD = 1,\: hyp = 2
    Hence: . AC \,=\,\sqrt{3}


    Draw diagonal BD.

    In right triangle BED\!:

    . . BD^2 \:=\:DE^2 + BE^2 \:=\:\left(\tfrac{5}{2}\right)^2 + \left(\tfrac{\sqrt{3}}{2}\right)^2 \:=\:\tfrac{25}{4} + \tfrac{3}{4} \:=\:\tfrac{28}{4} \:=\:7

    Therefore: . BD \:=\:\sqrt{7}

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  3. #3
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    Quote Originally Posted by Hanga View Post
    Okay I'll just ask the question.

    If I know that the sides of a parallellogram is 1 (l.u = length unit ) and 2 l.u and that one angle is 60degrees, can I using plain Euklides Geometry calculate the diagonals exactly?

    No need for you guys to solve it completly but a tip would be nice!

    I can get all angles easy. This would be a breeze if it was a rectangle, but cuz it's leaning the diagonal must become somewhat longer. I tried to get it by creating it into a 90degree traingle so I could use phytagoras, but for that I'd have to make a million small triangles and I hardly belive it's the easiest way.

    Thanks in advance.
    Here comes a different attempt:

    Use the Cosine rule in the triangle CDA to calculate the diagonal AC.

    Use the Cosine rule in the triangle DCB to calculate the diagonal DB.

    Since \cos(60^\circ) = 0.5 and \cos(120^\circ) = -0.5 the calculations are really simple.
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  4. #4
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    Thanks alot, both of you!

    The latter explenation I could've done, problem was that I had to proove it using euclides geometry only!

    I understand this now, thx alot
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