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Math Help - Intercept theorem in triangles

  1. #1
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    Intercept theorem in triangles

    Its straight line geometry time and I've reached another point that I cannot seem to get my head round (though I am sure it will be something obvious).

    The Question

    Triangles PQR and XYZ are such that:

    1. Angle P = Angle X
    2. Angle Q = Angle Z
    3. XY = 3cm
    4. YZ = 4cm
    5. PQ = 7cm
    6. PR = 12cm

    I need to find the lengths of XZ and QR.

    My answer so far

    So I have a large PQR triangle drawn on my page with a line ZY which is parallel to QR, the symbols for angles are put in there to indicate that the two triangles are similar and that PQR is an enlargement of XYZ.

    So to find XZ I knew the ratios of the intercepts would be the same so:

    PZ:QR = PY:PR

    so PZ:7 = 3:12 (or 1:4)

    so PZ = (7*3)/12 (or 7/4)

    The problem

    How do I go about working out QR?
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  2. #2
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    Draw triangle XYZ on PQR.
    P coincides X, Z lies on PQ and Y lies on PR.
    Since angle Z = angle Q and angle Y =angle R, ZY is parallel to QR
    The two triangles a similar. So
    XY/PR = XY/PQ = ZY/QR.
    Now find the required sides.
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  3. #3
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    Hello, Alice!

    You've done all the preliminary work.
    . . The answers are waiting for you . . .


    Triangles PQR and XYZ are such that:

    . . \begin{array}{ccc}\angle P &=& \angle X \\<br />
\angle Q &=& \angle Z \\ XY &=& 3\text{cm} \\ YZ &=& 4\text{cm}\\  PQ &=& 7\text{cm} \\ PR &=& 12\text{cm} \end{array}

    Find the lengths of XZ and QR.

    Did you make a sketch like this?


    Code:
                  P
                  *
                 *  *
                *     *
               *        *                     X
            7 *           * 12                *
             *              *                *  *
            *                 *             *     * 3
           *                    *          *        *
        Q *  *  *  *  *  *  *  *  * R   Z *  *  *  *  * Y
                                                4

    The two triangles are similar; their sides are proportional.


    Since PR = 12 and XY = 3,

    . . we see that \Delta PQR is four times as large as \Delta XYZ


    Then: . QR \:=\:4\cdot YZ \quad\Rightarrow\quad QR \:=\:4(4) \quad\Rightarrow\quad QR\:=\:16

    And: . PQ \:=\:4\cdot XZ \quad\Rightarrow\quad 7 \:=\:4\cdot XZ \quad\Rightarrow\quad XZ \:=\:\tfrac{7}{4}

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  4. #4
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    My thanks to both of you, it makes a lot more sense now =)
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  5. #5
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    Jerusalem - Israel
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    try to continue

    Quote Originally Posted by AliceFisher View Post
    Its straight line geometry time and I've reached another point that I cannot seem to get my head round (though I am sure it will be something obvious).

    The Question

    Triangles PQR and XYZ are such that:

    1. Angle P = Angle X
    2. Angle Q = Angle Z
    3. XY = 3cm
    4. YZ = 4cm
    5. PQ = 7cm
    6. PR = 12cm

    I need to find the lengths of XZ and QR.

    My answer so far

    So I have a large PQR triangle drawn on my page with a line ZY which is parallel to QR, the symbols for angles are put in there to indicate that the two triangles are similar and that PQR is an enlargement of XYZ.

    So to find XZ I knew the ratios of the intercepts would be the same so:

    PZ:QR = PY:PR

    so PZ:7 = 3:12 (or 1:4)

    so PZ = (7*3)/12 (or 7/4)

    The problem

    How do I go about working out QR?

    Attached Thumbnails Attached Thumbnails Intercept theorem in triangles-razem.jpg  
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  6. #6
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    So pretty
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