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Math Help - [SOLVED] parallelograms

  1. #1
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    [SOLVED] parallelograms

    The diagonals of a parallelogram have measures of 8 and 10 and intersect at a 60 degree angle. Find the area of the parallelogram.
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  2. #2
    Moo
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    Hello,

    See the sketch below...

    Hint : use trigonometry in triangle OAD. Everything else is there
    Attached Thumbnails Attached Thumbnails [SOLVED] parallelograms-trigpar.jpg  
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    thank you

    then that means the height of the triangle is 5√(3) and its base is 10. so the area is 25√(3) making the whole parallelogram's area to be 50√(3). i think that's right
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    Moo
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    Quote Originally Posted by majik92 View Post
    thank you

    then that means the height of the triangle is 5√(3) and its base is 10. so the area is 25√(3) making the whole parallelogram's area to be 50√(3). i think that's right
    Hmmm, there is a slight error

    How do you find 5√(3) ? Can you check it ?
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    oops, the height's 2√(3) -> A of triangle 5√(3) -> A of parallelogram 10√(3)
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    Moo
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    Quote Originally Posted by majik92 View Post
    oops, the height's 2√(3) -> A of triangle 5√(3) -> A of parallelogram 10√(3)
    Yes, that's better ^^
    But see the red part
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  7. #7
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    I notice that this thread is marked ‘solved’.
    But in fact it is not. There is no correct solution yet.
    What should have been noted from the beginning is: the area of the parallelogram is 4\cdot\mbox{area}(\Delta AOD).
    Now \mbox{area}(\Delta AOD) = \frac{1}{2}\left( 5 \right)\left( 4 \right)\sin \left( {60^ \circ  } \right).
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  8. #8
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    Quote Originally Posted by Plato View Post
    I notice that this thread is marked ‘solved’.
    But in fact it is not. There is no correct solution yet.
    What should have been noted from the beginning is: the area of the parallelogram is 4\cdot\mbox{area}(\Delta AOD).
    Now \mbox{area}(\Delta AOD) = \frac{1}{2}\left( 5 \right)\left( 4 \right)\sin \left( {60^ \circ  } \right).
    but i found the area of triangle DAB (as per Moo's graphic) and doubled it b/c triangle DAB is congruent to triangle BCD
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    Quote Originally Posted by majik92 View Post
    but i found the area of triangle DAB (as per Moo's graphic) and doubled it b/c triangle DAB is congruent to triangle BCD
    But what you did may be wrong.
    Go over your work.
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  10. #10
    Moo
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    \mathcal{A}_{ADB}=\frac 12 \cdot AH \cdot DB=\frac 12 (\cos 60 \cdot OA) \cdot DB

    \mathcal{A}_{ADB}=\frac 12 \cdot \frac{\sqrt{3}}{2} \cdot 4 \cdot 10=10\sqrt{3}

    Area of the parallelogram is twice this one...


    Same result as Plato's, slight different method... This was going to be the correct result ~
    Last edited by Moo; June 17th 2008 at 01:23 AM.
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