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Math Help - Help finding the length of a line?

  1. #1
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    Help finding the length of a line?

    Turns out that I either suck at algebra (likely) or I suck at algebra at 3:20 in the morning (also likely).

    In the diagram, Q lies on the joining (0 , 6) and (3 , 0).

    OPQR is a rectangle, where P and R lie on the axes and OR = length "t".

    (a) Show that QR = 6 - 2t.



    I've had a variety of ideas for this, but none of them seem to be quite right. Any help at all would be appreciated.
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  2. #2
    MHF Contributor harish21's Avatar
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    first of all, find the area of the large triangle.. Let OS be the base of the large triangle whose length is 3.

    Let OU be the perpendicular, and its given length is 6

    So, the area of the large triangle UOS is \dfrac{1}{2}\times \mbox{ base} \times \mbox{height} = \dfrac{1}{2}\times 6 \times 3=9.......(I)

    Now, the side OR of the rectangle has is OR=QP=t. and since OR=t, you have RS=3-t

    let the length of the other side of the rectangle be OP=x. therefore, PU=6-x

    and the area of the rectangle is tx...........(II)

    Now the area of the small triangle QRS = \dfrac{1}{2}\times \mbox{RS} \times \mbox{QR}........(III)

    and the area of the triangle QPU = \dfrac{1}{2}\times \mbox{PU} \times \mbox{PQ}.....(IV)

    now you should know that the area of triangle OPU+ the area of triangle QRS+ the area of the rectangle = The area of the triangle UOS...

    Use (I)=(II)+ (III)+ (IV) to find x...
    Last edited by harish21; January 26th 2011 at 07:29 PM.
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  3. #3
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    Thanks for the reply.

    I can follow you completely until I try and get the area of triangle QPU - I can't seem to figure out what PU would be equal to. You say that OU = 6-t, but I thought that OU was just equal to 6?

    Edit: would PU be 6 (length of OU) - x (6-2t)?
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  4. #4
    MHF Contributor harish21's Avatar
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    Quote Originally Posted by mtndew View Post
    Thanks for the reply.


    I can follow you completely until I try and get the area of triangle QPU - I can't seem to figure out what PU would be equal to. You say that OU = 6-t, but I thought that OU was just equal to 6?
    sorry that was a typo.. I corrected it..

    OU is the height of the traingle and its length is 6.

    You are interested in finding OP, which is equal to QR. so OP=QR= x.

    so, PU= OU-OP = 6-x

    the area of triangle QPU = \dfrac{1}{2} \times PU \times PQ = \dfrac{1}{2} \times (6-x) \times t
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  5. #5
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    Hello, mtndew!

    In the diagram, \,Q lies on the joining B(0,6) and A(3,0).

    OPQR is a rectangle, where \,P and \,R lie on the axes and OR = t.

    (a) Show that: . QR \:=\:6 - 2t

    Code:
            |
        - B *
        :   | *
        :   |   *
        :   |     *
        :   |       *
        6   |         *
        : P * - - - - - * Q
        :   |           | *
        :   |           |   *
        :   |           |     *
        - - + - - - - - * - - - *
            O     t     R  3-t  A
            : - - - - 3 - - - - :

    Since \Delta QRA \sim \Delta BOA,\;\dfrac{QR}{3-t} \:=\:\dfrac{6}{3} \quad\Rightarrow\quad QR \:=\:6 - 2t

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