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Math Help - rhombus problem

  1. #1
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    rhombus problem

    Use the converse rule that the diagonals of a rhombus bisect each other at 90 degrees and that when 2 lines bisect each other at 90 degrees the quadrilateral formed by joining the points is a rhombus to show:
    that the points with co-ordinates (1,2) (8,-2) (7,6) (0,10) are the vertices of a rhombus, and find the area.

    Any pointers?
    I am thinking along the lines of proving that the angles in the middle are 90, but not sure how to go about it, I thought about using pythag but how would I find the length of the diagnonals?
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  2. #2
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    Quote Originally Posted by Sweeties View Post
    Use the converse rule that the diagonals of a rhombus bisect each other at 90 degrees and that when 2 lines bisect each other at 90 degrees the quadrilateral formed by joining the points is a rhombus to show:
    that the points with co-ordinates (1,2) (8,-2) (7,6) (0,10) are the vertices of a rhombus, and find the area.

    Any pointers?
    I am thinking along the lines of proving that the angles in the middle are 90, but not sure how to go about it, I thought about using pythag but how would I find the length of the diagnonals?
    To start: Draw a sketch!

    (1, 2) and 7,6)
    (0,10) and (8,-2) are the opposite vertices of the rhombus.

    If the diagonals are perpendicular then the product of the slopes must be (-1)

    Check if the midpoint in both cases is the same.

    The area of the rhombus is calculated by:

    A=\frac12 \cdot (\text{length of the 1rt diagonal}) \cdot (\text{length of the 2nd diagonal})

    The distance between 2 points P_1(x_1, y_1) and P_2(x_2, y_2) is calculated by:

    d = \sqrt{(x_2-x_1)^2+(y_1-y_2)^2}
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  3. #3
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    Quote Originally Posted by earboth View Post
    To start: Draw a sketch!

    (1, 2) and 7,6)
    (0,10) and (8,-2) are the opposite vertices of the rhombus.

    If the diagonals are perpendicular then the product of the slopes must be (-1)

    Check if the midpoint in both cases is the same.

    The area of the rhombus is calculated by:

    A=\frac12 \cdot (\text{length of the 1rt diagonal}) \cdot (\text{length of the 2nd diagonal})
    Thanks, I did the sketch already! OK, so if I find the gradient of each diagonal and the product of both is -1 then they are pependicular and so the angles must be 90 degrees. Got it!
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  4. #4
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    Quote Originally Posted by Sweeties View Post
    Thanks, I did the sketch already! OK, so if I find the gradient of each diagonal and the product of both is -1 then they are pependicular and so the angles must be 90 degrees. Got it!
    You certainly know that this property isn't sufficient to determine a rhombus (see attachment: The diagonals are perpendicular but the quadrilateral is not a rhombus!)

    The midpoint between 2 points P_1(x_1, y_1) and P_2(x_2, y_2) is calculated by:

    M\left(\frac{x_1+x_2}{2} , \frac{y_1+y_2}{2}\right)
    Attached Thumbnails Attached Thumbnails rhombus problem-kein_rhombus.gif  
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  5. #5
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    Quote Originally Posted by earboth View Post
    You certainly know that this property isn't sufficient to determine a rhombus (see attachment: The diagonals are perpendicular but the quadrilateral is not a rhombus!)

    The midpoint between 2 points P_1(x_1, y_1) and P_2(x_2, y_2) is calculated by:

    M\left(\frac{x_1+x_2}{2} , \frac{y_1+y_2}{2}\right)
    I worked out the midpoints, they are equal.
    The centre is (4,4)
    the smaller diagonal is 6 in lenght so half is 3
    the longer diagonal is 8 in length so half is 4

    So the perpendicular rule will work for this rhombus whereas is wouldn't have for the one you've drawn. OUt of curiosity, how would it work with the one you have drawn?
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  6. #6
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    Quote Originally Posted by Sweeties View Post
    I worked out the midpoints, they are equal.
    The centre is (4,4)
    the smaller diagonal is 6 in lenght so half is 3
    the longer diagonal is 8 in length so half is 4

    So the perpendicular rule will work for this rhombus whereas is wouldn't have for the one you've drawn. OUt of curiosity, how would it work with the one you have drawn?
    You get all the same properties with 3 exceptions:

    1. The diagonals don't have a common midpoint.

    2. The sides of the quadrilateral (in German such a figure is called ein Drachen = a kite(?); I don't know how you call it in English) have different length. (Remember: The sides of a rhombus have equal length)

    3. Opposite angles don't have the same size.


    By the way: The area of your rhombus is 24.
    Last edited by earboth; February 9th 2008 at 12:28 AM.
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    rhombus problem

    i went over your problem and think you have t...he wrong lenghts of the diagonals.slope triangle legs are 8 and 12, 4and 6. long diagonal is the square root of12 squared plus 8 squared. short diagonal is square root of4squared plus6squared.
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  8. #8
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    Quote Originally Posted by bjhopper View Post
    i went over your problem and think you have t...he wrong lenghts of the diagonals.slope triangle legs are 8 and 12, 4and 6. long diagonal is the square root of12 squared plus 8 squared. short diagonal is square root of4squared plus6squared.
    Hi,

    thanks for your reply.
    I did't check these results:
    Quote Originally Posted by Sweeties View Post
    I worked out the midpoints, they are equal.
    The centre is (4,4)
    the smaller diagonal is 6 in lenght so half is 3
    the longer diagonal is 8 in length so half is 4
    As bjhopper pointed out the length of the diagonals are:

    d_1 = \sqrt{8^2+2^2} = \sqrt{208} = 2 \cdot \sqrt{52}

    d_2 = \sqrt{6^2+4^2} = \sqrt{52}

    And therefore the area is:

    a = \frac12 \cdot 2 \cdot \sqrt{52} \cdot \sqrt{52} = 52
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