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Math Help - Trapezium vector problem

  1. #1
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    Trapezium vector problem

    ABCD is a trapezium, AB and CD are parallel sides, M is the midpoint of AD. Prove that the area of the triangle BMC is half the area of the trapezium.

    I let A be the origin, then the vectors of AB, AC, and AD respectively are b, c, d. So vector AM= \frac{1}{2}d
    Area of the trapezium= \frac{1}{2}|b\times c|+\frac{1}{2}|c\times d|
    vector CM= \frac{1}{2}d-c vector CB= b-c
    Area BMC= \frac{1}{2}|(b-c)\times (\frac{1}{2}d-c)|
    My problem is I am stuck here, I think it is just some rearrangement of the vectors, but I am no good at dot and cross-products.
    Thanks
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  2. #2
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    Quote Originally Posted by arze View Post
    ABCD is a trapezium, AB and CD are parallel sides, M is the midpoint of AD. Prove that the area of the triangle BMC is half the area of the trapezium.

    I let A be the origin, then the vectors of AB, AC, and AD respectively are b, c, d. So vector AM= \frac{1}{2}d
    Area of the trapezium= \frac{1}{2}|b\times c|+\frac{1}{2}|c\times d|
    vector CM= \frac{1}{2}d-c vector CB= b-c
    Area BMC= \frac{1}{2}|(b-c)\times (\frac{1}{2}d-c)|
    My problem is I am stuck here, I think it is just some rearrangement of the vectors, but I am no good at dot and cross-products.
    Thanks
    Is it necessary to use vectors? If not:

    1. The area of a trapezium is calculated by:

    A = \frac12 \cdot (a+c) \cdot h

    2. With h_a = h_c = \frac12 h you can calculate the areas of the triangles with the bases a or c respectively. Subtract these areas from the area of the trapezium and you'll get the area of the triangle BMC:

    3. A_{\Delta(BMC)}= \frac12 \cdot (a+c) \cdot h - \frac12 \cdot a \cdot h_a - \frac12 \cdot c \cdot h_c

    A_{\Delta(BMC)}= \frac12 \cdot (a+c) \cdot h - \frac12 \cdot a \cdot \frac12 \cdot h - \frac12 \cdot c \cdot \frac12 \cdot h

    4. Expand and collect like terms. You'll get A_{\Delta(BMC)}= \frac14 \cdot (a+c) \cdot h which is half of the trapezium's area.
    Attached Thumbnails Attached Thumbnails Trapezium vector problem-trapez_haelfte.png  
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  3. #3
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    Yes, I am supposed to use vectors.
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    Refer the attached image in#2. Draw MN parallel to AB.

    MN = \frac{a+c}{2}

    Let the vector CB be b.

    Now the area of the triangle MNC = \frac{1}{2}(\frac{a+c}{2}\times\frac{b}{2})

    Area of the triangle BMC = 2*MNC = \frac{1}{2}(\frac{a+c}{2}\times{b})

    = \frac{1}{2}(\frac{{a}\times{b}}{2} + \frac{{c}\times{b}}{2})

    = 1/2*[ares(ABC) + ares(DBC)]

    But Area(ABC) = area(ADC)

    So Area(BMC) = 1/2*[area(ADC) + ares(DBC)] = 1/2[areaABCD]
    Last edited by sa-ri-ga-ma; June 30th 2010 at 05:26 AM.
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  5. #5
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    What is N?
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  6. #6
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    Quote Originally Posted by arze View Post
    What is N?
    N is the midpoint of BC.
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