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Math Help - Area of Parallelograms

  1. #1
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    Area of Parallelograms

    find the perimeter and area of each parallelogram Round to the nearest tenth if necessary



    figure is exactly like that
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    Quote Originally Posted by Niceguy78 View Post
    find the perimeter and area of each parallelogram Round to the nearest tenth if necessary



    figure is exactly like that
    note the 30-60-90 triangle formed in the upper part of the parallelogram ... the altitude of the parallelogram is the long leg of the 30-60-90 triangle.

    ratios of the sides of a 30-60-90 triangle ...

    short leg = x
    long leg = x\sqrt{3}
    hypotenuse = 2x

    finally, area of a parallelogram is A = bh
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    im looking for the area AND parallelogram and i dont get how you get the x\sqrt{3}
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    Quote Originally Posted by Niceguy78 View Post
    im looking for the area AND parallelogram and i dont get how you get the x\sqrt{3}
    If you have no knowledge of the properties of a 30-60-90 right triangle, then you shouldn't have been assigned this problem.

    Go to this site and learn about them ...

    30 60 90 Right Triangles - Free Math Help
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    ok ty but now all the problem that i have here are 30 60 90 righ triangles some contain different measurements
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    hi i kind of get it now i got the answer 780 anyone can tell me if i got it right?? sorry for double post
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    It is fairly well known that the area of a parallelogram is A=a\cdot b\cdot \sin(\theta) there a~\&~b are the lengths of two adjacent sides and \theta is the angle between then sides.
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    o wow confusing me.. you give me a different formula?? is my answer right?
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    Quote Originally Posted by Niceguy78 View Post
    o wow confusing me.. you give me a different formula?? is my answer right?
    The area of your parallelogram is (20)(30)\sin\left(60^0\right).
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