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Math Help - Diagram problem

  1. #1
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    Diagram problem

    The diagram consists of a rectangle ABCD and a triangle DXY so that X and Y are points on the line segments AB and BC respectively and angle DXY=90.
    Show that DA/XB =AX/BY =XD/YX

    Thanks guys.
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  2. #2
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    Hello, hongvo!

    The diagram consists of a rectangle ABCD and a triangle DXY
    with X and Y on AB and BC, resp. and \angle DXY\,=\,90^o

    Show that: . \frac{DA}{XB} \:=\: \frac{AX}{BY} \:=\: \frac{XD}{XY}
    Code:
                    X
        A o - - - - o - - - - - - - - - o B
          |      β *    *   α           |
          |       *  90    *           |
          |      *              *       |
          |     *                   * β |
          |    *                        o Y
          |   *                         |
          |α *                          |
          | *                           |
          |*                            |
        D o - - - - - - - - - - - - - - o C

    \text{In right triangle }XAD\!: \begin{array}{ccccccc}\text{let}& \alpha &=& \angle ADX \\<br /> <br />
\text{let} & \beta &=& \angle AXD &=& 90^o-\alpha \end{array}


    \text{At }X\!:\;\;\angle AXD + 90^o + \angle BXY \:=\:180^o \quad\Rightarrow\quad \beta + 90^o + \angle BXY \:=\:180^o


    And we have: . \begin{array}{ccccc}\angle BXY &=& 90^o - \beta &=& \alpha \\<br />
\angle BYX &=& 90^o - \alpha &=& \beta \end{array}


    Hence: . \Delta XAD \:\sim\:\Delta YBX


    Therefore: . \frac{DA}{XB} \;=\;\frac{AX}{BY} \;=\;\frac{XD}{XY}

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  3. #3
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    Smile diagram problem

    Hi Soroban

    Thank you very much for your help. All the best n have a good day.

    Cheers, hv
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  4. #4
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    sophie diagram problem

    Can anyone please tell me if they have solved part c of this question which is:
    c. Another sophie diagram has AX=15, YC=16 and XB=48. Find the lengths of the sides of triangle DXY

    thanks
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  5. #5
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    Hello, imcnalty!

    Who is "sophie"?


    c. Another sophie diagram has: .  AX\,=\,15,\;\;YC\,=\,16,\;\;XB\,=\,48.

    Find the lengths of the sides of \Delta DXY.
    Code:
              15    X        48
        A o - - - - o - - - - - - - - - o B
          |        *    *               |
          |       *  90    *           | b
          |      *              *       |
        a |     *                   * β |
          |    *                        o Y
          |   *                         |
          |α *                          | 16
          | *                           |
          |*                            |
        D o - - - - - - - - - - - - - - o C
                        63

    Draw DY.\quad DC = 63.

    Let: . a \,=\,AD,\;\;b\,=\,BY.

    Note that: . a \:=\:b+165 .[1]


    In right triangle XAD\!:\;\;DX^2 \:=\:a^2+15^2 .[2]

    In right triangle XBY\!:\;\;XY^2 \:=\:b^2 + 48^2 .[3]

    In right triangle YCD\!:\;\;DY^2 \:=\:16^2 + 63^2 \:=\:4225 .[4]


    In right triangle DXY\!:\;\;DX^2 + XY^2 \:=\:DY^2

    Substitute [2], [3], [4]: . (a^2+15^2) + (b^2 + 48^2) \:=\:4225  \quad\Rightarrow\quad a^2+b^2 \:=\:1696

    Substitute [1]: . (b+16)^2 + b^2 \:=\:1696 \quad\Rightarrow\quad 2b^2 + 32b - 1440 \:=\:0

    . . . . . . . . . . . 2(b-20)(b+36) \:=\:0 \quad\Rightarrow\quad b \:=\:20 \quad\Rightarrow\quad a \:=\:36


    Therefore: . \begin{Bmatrix}DX &=&\sqrt{36^2+15^2} &=& 39 \\<br />
XY &=& \sqrt{20^2+48^2} &=& 52 \\<br />
DY &=& \sqrt{4225} &=& 65 \end{Bmatrix}

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  6. #6
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    sophie diagram

    hi can you help me with this one to??

    DX= 429
    DY= 845!!
    hurry please
    thankyou soooo muchhh xxxx
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  7. #7
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    diagram prob

    Hi Soroban

    I've used the Pythagoras triangle (Triads) to get the answers. Is it still correct? The answers are the same in the end.

    Cheers, hv
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  8. #8
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    Sup katepark , are you in year 9 or 10? I suppose you're doing the Maths Challenge too yeah?
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