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Thread: Area of the quadrilateral

  1. #1
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    Area of the quadrilateral

    Area of the quadrilateral-2010813174745_0.jpg
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  2. #2
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    The triangles BOC, OQC, on the same base BQ, have the same "height" (namely the perpendicular distance from C to BQ). Since the area of BOC is twice the area of OQC, it follows that BO = 2OQ. By a similar argument from the triangles COB, OBP, it follows that CO = \frac54OP.

    Now let x be the area of triangle PAO, and let y be the area of triangle OAQ. Since BO = 2OQ, it follows (by the same reasoning as in the previous paragraph) that the area of triangle BAO is twice the area of triangle OAQ. In other words, 8+x = 2y. A similar argument in triangles PAO and OAC shows that 5+y = \frac54x.

    That gives you two simultaneous equations for x and y. The answer to the problem is the sum x+y.
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    Thank you again Opalg, realy it is a logical solution.
    The answer to the problem is the sum x+y as you say
    x = 12 , y = 10
    and x + y = 22
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  4. #4
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    Quote Originally Posted by razemsoft21 View Post
    Click image for larger version. 

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    Alternatively, you can use ratios...

    Using [BQ]

    x+8)" alt="5:10=yx+8)" />

    Using [PC]

    (5+y):x=10:8



    This is because......


    h_1=perpendicular\ height\ of\ triangles\ OAB\ and\ OAQ

    \displaystyle\huge\frac{1}{2}|OQ|h_1=y,\ \frac{1}{2}|OB|h_1=x+8\ \Rightarrow\frac{h_1}{2}=\frac{y}{|OQ|}=\frac{x+8}  {|OB|}\Rightarrow\frac{|OB|}{|OQ|}=\frac{x+8}{y}


    h_2=perpendicular\ height\ of\ triangles\ OQC\ and\ OBC

    \displaystyle\huge\frac{1}{2}|OQ|h_2=5,\ \frac{1}{2}|OB|h_2=10\Rightarrow\frac{h_2}{2}=\fra  c{5}{|OQ|}=\frac{10}{|OB|}\Rightarrow\frac{|OB|}{|  OQ|}=\frac{10}{5}


    h_3=perpendicular\ height\ of\ triangles\ OAP\ and\ OAC

    \displaystyle\huge\frac{1}{2}|OP|h_3=x,\ \frac{1}{2}|OC|h_3=y+5\Rightarrow\frac{h_3}{2}=\fr  ac{x}{|OP|}=\frac{y+5}{|OC|}\Rightarrow\frac{|OC}{  OP|}=\frac{y+5}{x}


    h_4=perpendicular\ height\ of\ triangles\ OBP\ and\ OBC

    \displaystyle\huge\frac{1}{2}|OP|h_4=8,\ \frac{1}{2}|OC|h_4=10\Rightarrow\frac{h_4}{2}=\fra  c{8}{|OP|}=\frac{10}{|OC|}\Rightarrow\frac{|OC|}{|  OP|}=\frac{10}{8}
    Last edited by Archie Meade; Aug 16th 2010 at 03:14 AM.
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