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Thread: Area of a Parallelogram

  1. #1
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    Area of a Parallelogram

    Area of a Parallelogram-whatsapp-image-2017-08-21-2.43.44-pm.jpeg
    I have to find the area of this parallelogram, but the only three things I'm told are that AD = 5, EF = 11, and EFC = 90. So far I've tried using the pythagoras theorem, and the special 30-60-90 triangle theorem (longer cathet = √3 * shorter cather OR (√3/2) * Hypotenuse), but they all lead to a dead end.
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  2. #2
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    Re: Area of a Parallelogram

    $A = bh = 5 \cdot 11$
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    Re: Area of a Parallelogram

    But 11 is not the length, DC is the length, 11 is EF.
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  4. #4
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    Re: Area of a Parallelogram

    Quote Originally Posted by kaperdomo View Post
    But 11 is not the length, DC is the length, 11 is EF.
    base & height are perpendicular ...

    EF is the height

    AD = BC is the base
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    Re: Area of a Parallelogram

    Quote Originally Posted by skeeter View Post
    base & height are perpendicular ...

    EF is the height
    How is EF the height? DC is the height and nothing says they are congruent.

    Quote Originally Posted by skeeter View Post
    AD = BC is the base
    I get this, it's the EF part I have a problem with.
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  6. #6
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    Re: Area of a Parallelogram

    DC is not the height.

    One more time ... base and height of a parallelogram are perpendicular.

    turn the attached diagram 90 degrees ...

    Last edited by skeeter; Aug 21st 2017 at 03:00 PM.
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