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Math Help - geometry length

  1. #1
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    geometry length

    how else can i say this? the way i did it he doesnt want.
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  2. #2
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     FB = DB

     \sqrt{(FB)^2+(DB)^2} = \sqrt{8^2 \cdot 2}

     \sqrt{(DB)^2+(DB)^2} = 8 \sqrt{2}

     \sqrt{2(DB)^2} = 8 \sqrt{2}

     \sqrt{2}(DB)^2 = 8 \sqrt{2}

     DB = 8 = FB

    for the total area of segment AB we have 11+8 = 19 , therefore the area of the rectangle ANDB is 19*8 = 152 and the area of the triangle FDB is (8*8)/2 = 32 = ANG

    so subtracting 152 by 2(32) = 88 which is the area of the parallelogram

    now for the rectangle we have 11 * 8 = 88
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