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Thread: 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; Jan 30th 2010 at 07:41 AM. Reason: forgot to attach the figure..!
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  2. #2
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    Hello, snigdha!

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

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

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

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

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

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

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


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



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

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

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


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

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