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Math Help - Please help me with this simliarity sum..

  1. #1
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    Please help me with this simliarity sum..

    ABCD is a //gm, P is a point on BC such that BP:PC=1:2 and DP produced meets AB produced at Q. If area of △CPQ = 20 sq.cm, find:
    1. area of △BPQ
    2. area of △CPD
    3. area of //gm ABCD.


    pls. open the attached word document to view the figure...
    Attached Files Attached Files
    Last edited by snigdha; January 30th 2010 at 07:41 AM. Reason: forgot to attach the figure..!
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  2. #2
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    Hello, snigdha!

    ABCD is a parallelogram.
    Point P is on BC such that: . BP:PC\,=\,1:2
    DP produced meets AB produced at Q.
    If area of \Delta CPQ = 20 \text{ cm}^2, find:

    1. area of \Delta BPQ
    2. area of \Delta CPD
    3. area of parallelogram ABCD
    Code:
                D                   R    C
                o - - - - - - - - - + - -o
               /    *               :   /
              /         *           :  /
             /              *       : /
            /                   *   :/
           /                        o P
          /                        /:   *
         /                        / :       *
        o - - - - - - - - - - - -o- + - - - - - o
        A                        B  S           Q

    Draw line segment CQ.
    . . Area of \Delta CPQ\,=\,20
    Draw altitude RS through point P\!:\;\;h = RS.

    Since BP:PC = 1:2, then: . SP:PR = 1:2
    . . That is: . SP = \tfrac{1}{3}h,\;PR = \tfrac{2}{3}h

    Area of \Delta BCQ \,=\,\tfrac{1}{2}(BQ)(h)
    Area of \Delta BPQ \,=\,\tfrac{1}{2}(BQ)\left(\tfrac{1}{3}h\right)\,=  \,\tfrac{1}{6}(BQ)(h)
    . . Hence: . \Delta BPQ \:=\:\tfrac{1}{3}\Delta BCQ

    Then: . \Delta CPQ \,=\,\tfrac{2}{3}\Delta BCQ \quad\Rightarrow\quad 20 \,=\,\tfrac{2}{3}\Delta BCQ

    Hence: . \Delta BCQ\,=\,30 \quad\Rightarrow\quad \Delta BPQ \,=\,10 .(1)


    We see that: . \Delta CPD \sim \Delta BPQ
    . . and \Delta CPD has twice the base and twice the height of \Delta BPQ
    Hence: . \Delta CPD \:=\:4\times\Delta BPQ \:=\:4(10) \quad\Rightarrow\quad \Delta CPD\:=\:40 .(2)



    The area of \Delta CPD is: . \tfrac{1}{2}(CD)\left(\tfrac{2}{3}h\right) \:=\:40 \quad\Rightarrow\quad \tfrac{1}{3}(CD)(h) \:=\:40

    Hence:. . (CD)(h) \:=\:120

    Note that: . (CD)(h) = area of parallelgram ABCD.


    Therefore: .area of ABCD \:=\:120 .(3)

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