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Math Help - 2 problems

  1. #1
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    2 problems

    Hi, here are two problems i am still having trouble with today.

    1.1.ABCDEF is a regular hexagon and EDGH is a square. Find the measures of <EFH, <FHD and <DFH.

    IDK if this is right but i found this:Angle EDH = 45 angle EHD=45 Ang. HDG = 45 angle DHG=4 all angles in that square are right angles as well




    8.Two vertical poles have heights of 6 feet and 12 feet. A rope is stretched from the top of each pole to the bottom of the other. How far above the ground do the ropes cross? (hint: the lengths of y and z do not affect the answer.)
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  2. #2
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    [quote=abc123;253849]Hi, here are two problems i am still having trouble with today.

    1.1.ABCDEF is a regular hexagon and EDGH is a square. Find the measures of <EFH, <FHD and <DFH.

    IDK if this is right but i found this:Angle EDH = 45 angle EHD=45 Ang. HDG = 45 angle DHG=4 all angles in that square are right angles as well



    Find each interior angle of hexagon = \frac{180(n-2)}{n} = \frac{180(6-2)}{6} = 120

    DEF = 120,

    DEH = 90 (angles of a square.)

    EHD = 45 = EDH

    FE = EH ,so, EFH = EHF

    FEH = 360 - 120 -90 = 150

    In triangle EFH,

    EFH = 15 = EHF

    IN triangle FED,

    EF = ED

    EFD = EDF = 30

    so,

    FDH = 30 + 45 = 75

    DFH = 30 + 15 = 45
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  3. #3
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    Quote Originally Posted by abc123 View Post
    Hi, here are two problems i am still having trouble with today.

    8.Two vertical poles have heights of 6 feet and 12 feet. A rope is stretched from the top of each pole to the bottom of the other. How far above the ground do the ropes cross? (hint: the lengths of y and z do not affect the answer.)
    Use similar triangles properties,

    \frac{x}{6} = \frac{z}{y+z}

    and

    \frac{x}{12} = \frac{y}{y+z}

    add these two eqns,

    \frac{x}{6}+\frac{x}{12} = \frac{z}{y+z}+\frac{y}{y+z} = \frac{y+z}{y+z} = 1

    x = 4
    Last edited by Shyam; January 25th 2009 at 01:11 PM.
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  4. #4
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    Thanks a lot for the posts. they helped a lot. I'm still confused though on number 6. I don't really understand what you did at all. So, if you could explain it to me a bit I'd be very greatful
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  5. #5
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    Hello again, abc123!

    8. Two vertical poles have heights of 6 feet and 12 feet.
    A rope is stretched from the top of each pole to the bottom of the other.
    How far above the ground do the ropes cross?
    Code:
                                  * C
                                * |
                              *   |
                            *     |
                          *       |
                        *         |
        A *           *           | 12
          |   *   E *             |
          |       *               |
        6 |     * :   *           |
          |   *   :x      *       |
          | *     :           *   |
        B *-------*---------------* D
          : - y - F - - - z - - - :

    \Delta EFD \sim \Delta ABD \quad\Rightarrow\quad \frac{x}{z} \:=\:\frac{6}{y+z} \quad\Rightarrow\quad x(y+z) \:=\:6z\;\;{\color{blue}[1]}

    \Delta EFB \sim \Delta CDB \quad\Rightarrow\quad \frac{x}{y} \:=\:\frac{12}{y+z} \quad\Rightarrow\quad x(y+z) \:=\:12y\;\;{\color{blue}[2]}


    Equate [1] and [2]: . 6z \:=\:12y \quad\Rightarrow\quad z \:=\:2y


    Substitute into [2]: . x(y + 2y) \:=\:12y \quad\Rightarrow\quad x(3y) \:=\:12y \quad\Rightarrow\quad \boxed{x \:=\:4}

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  6. #6
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    Hello, abc123!

    1) ABCD{E}F is a regular hexagon and EDGH is a square.
    Find the measures of: \angle EFH,\;\angle FHD,\;\angle DFH
    Code:
            A * - - - * F
             /        :\*
            /         : \ *
           /          :  \  *
        B * 120      : E *   *
           \          :  /90 * *
            \         : /         * H
             \        :/         /
            C * - - - *         /
                      D   *    /
                              *
                              G

    All interior angles of the hexagon are 120.

    We have: . \angle FED = 120^o,\;\angle DEH = 90^o
    . . Hence: . \angle FEH = 150^o

    \Delta EFH is isosceles, EF = EH
    . . Therefore: . \angle EFH \:=\:\angle EHF \:=\:15^o


    In square EDGH,\;\angle EHD = 45^o
    . . Therefore: . \angle FHD \:=\:15^o + 45^o \:=\:60^o


    \angle AFE = 120^o,\;\angle AFD = 90^o\quad\Rightarrow\quad \angle DFE = 30^o
    . . Therefore: . \angle DFH \:=\:30^o + 15^o \:=\:45^o

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