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Math Help - a few math team problems need help with

  1. #1
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    a few math team problems need help with

    1. from a point within an equilateral triangle, perpendiculars of length 3,6,9 are dropped to the sides. find the area of the triangle.

    2. What is the distance between the incenter and circumcenter of a triangle with sides of the length 12, 16 and 20 (*sidenote* 3,4,5 triangle right there)

    3. A right triangle's hypotenuse has projections of length 4 and 9. find the area of the triangle.

    4. A 6ft. tall man is standing 3ft away from a 9ft tall light source. How long is his shadow



    You don't have to answer all of them. answer the ones you can and provide a solution i will really appreciate it thank you!

    -Amer
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  2. #2
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    Quote Originally Posted by amerlaw1 View Post
    4. A 6ft. tall man is standing 3ft away from a 9ft tall light source. How long is his shadow
    Best thing to do is draw a picture to truly understand this one.

    Using similar triangles and equating ratios of side lengths you should get

    \displaystyle \frac{x}{6} = \frac{x+3}{9}

    Can you find x ?
    Last edited by pickslides; December 23rd 2010 at 12:59 PM. Reason: spelling mistakes
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  3. #3
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    Quote Originally Posted by amerlaw1 View Post
    1. from a point within an equilateral triangle, perpendiculars of length 3,6,9 are dropped to the sides. find the area of the triangle.

    2. What is the distance between the incenter and circumcenter of a triangle with sides of the length 12, 16 and 20 (*sidenote* 3,4,5 triangle right there)

    3. A right triangle's hypotenuse has projections of length 4 and 9. find the area of the triangle.

    4. A 6ft. tall man is standing 3ft away from a 9ft tall light source. How long is his shadow



    You don't have to answer all of them. answer the ones you can and provide a solution i will really appreciate it thank you!

    -Amer
    What is "mathteam"? Are these questions part of a competition?
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  4. #4
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    thank you but can u show me how the are similar like what theorem. E.G sas or aa or sss????

    yes math team is a competition but these questions came from a competition that happened 14 years ago so you need not to worry about cheating
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  5. #5
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    1. from a point within an equilateral triangle, perpendiculars of length 3,6,9 are dropped to the sides. find the area of the triangle.

    Let the side of the triangle be a. Then you can find the areas of AOB, BOC and AOC, which together give the area of ABC. On the other hand, the area of ABC is \sqrt{3}a^2/4. This gives you an equation that allows finding a.
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  6. #6
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    Hello, amerlaw1!

    2. What is the distance between the incenter and circumcenter
    of a triangle with sides of the length 12, 16 and 20?
    Code:
      _ A *
      :   | *
      :   |   *
      :   |     *
      :   |       *
      :   |         *
      :   |           *   20
      :   |       * * * *
      :   |   *           *   E
     12   | *               o
      :   |*          r   *  *
      :   |             *       *
      :   *           *       *   * 
      : F o - - - - o         *     *
      :   *    r    |O        *       *
      :   |         |                   *
      :  r|*        |r       *            *
      :   | *       |       *               *
      :   |   *     |     *                   *
      - C * - - - * o * - - - - - - - - - - - - * B
              r     D
          : - - - - - - - - 16  - - - - - - - - :

    We have right triangle ABC:\:AC = 12,\;BC = 16,\;AB = 20.
    Place the triangle on a coordinate system with \,C at the Origin.

    The incenter is O(r,r).
    . . OD = OE = OF = CD = CF = r.

    The circumcenter is the midpoint of AB:\;P(8,6).
    . .
    (Not shown on the diagram.)

    Note that: . AF \,=\, 12-r
    Since AE is also tangent to the circle:  AE \,=\, 12-r

    Note that: . BD \,=\,16-r.
    Since BE is also tangent to the circle: BE \,=\,16-r

    Since AB = 20, we have: . (12-r) + (16-r) \:=\:20 \quad\Rightarrow\quad r \,=\,4


    We have: . \begin{Bmatrix}\text{Incenter:} & O(4,4) \\ \text{Circumcenter:} & P(8,6) \end{Bmatrix}


    Therefore: . \iverline{OP} \;=\;\sqrt{(8-4)^2 + (6-5)^2} \;=\;\sqrt{16+4} \;=\;\sqrt{20} \;=\;2\sqrt{5}

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