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Math Help - Cone

  1. #1
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    Cone

    A right circular cone is inscribed in a regular tetrahedron whose edge is 10cm.Find the volume of the cone. Help please.
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  2. #2
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    Hello, reiward!

    This one takes a lot of preliminary work . . .


    A right circular cone is inscribed in a regular tetrahedron with edge 10 cm.
    Find the volume of the cone.
    Consider the base of the tetrahedron (and the cone).

    Code:
                          A
                          *
                         /|\
                        / | \
                       /  |  \
                      /   |   \
                     /    |    \
                    /   * * *   \
                   /*     |     *\
               10 *       |       * 10
                 *        |        *
                /         |         \
               /*         |         *\
              / *       O *         * \
             /  *         |  *      *  \
            /             | 60 *       \
           /     *        |        *     \
          /       *       |       *   *   \
         /          *     |     *    30 * \
      B *---------------*-*-*---------------* C
        : - - -  5  - - - D - - -  5  - - - :

    The base of the tetrahedron is equilateral triangle ABC,
    . . with: . AB = BC = CA = 10.
    AD is an altitude of the triangle: . BD =  DC = 5

    The base of the cone is the inscribed circle with center O.
    . . Its radius is OD.

    We find that \Delta ODC is a 30-60 right triangle with  DC = 5.
    . . Hence: . OD \,=\,\frac{5}{\sqrt{3}},\;OC \,=\,\frac{10}{\sqrt{3}}

    The radius of the cone is: . \boxed{r \,=\,\frac{5}{\sqrt{3}}}



    Consider a vertical cross-section of the tetrahedron,
    . . cut through the top vertex V and OC.
    Code:
        V *
          |\
          | \
          |  \
          |   \
        h |    \ 10
          |     \
          |      \
          |       \
          |        \
        O *---------* C
                 _
             10/√3

    We have: . h^2 + \left(\frac{10}{\sqrt{3}}\right)^2 \:=\:10^2 \quad\Rightarrow\quad h^2 + \frac{100}{3} \:=\:100 \quad\Rightarrow\quad h^2 \:=\:\frac{200}{3}

    The height of the cone is: . \boxed{h \:=\:\frac{10\sqrt{6}}{3}}


    Therefore, the volume of the cone is:

    . . . V \;=\;\frac{\pi}{3}r^2h \;=\;\frac{\pi}{3}\left(\frac{5}{\sqrt{3}}\right)^  2\left(\frac{10\sqrt{6}}{3}\right) \;=\;\frac{250\pi\sqrt{6}}{27} \text{ cm}^3

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