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Math Help - Proof

  1. #1
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    Proof

    Given a circle with radius t, and an inscribed regular hexagon; prove that the area of hexagon is a mean proportional between the areas of the inscribed and circumscribed equilateral triangles. Urgent help please!
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  2. #2
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    Hello, xoluosox!

    An equilateral triangle has area: . A \:=\:\frac{\sqrt{3}}{4}s^2
    . . where s is a side of the triangle.


    Given a circle with radius r, and an inscribed regular hexagon.
    Prove that the area of hexagon is a mean proportional between the areas
    of the inscribed and circumscribed equilateral triangles.
    Code:
                  * o *
              *     o   o *
            *     r o       o
           *        o     o  *
                    o   o r
          *         o o       *
          *         *         *
          *                   *
    
           *                 *
            *               *
              *           *
                  * * *
    The inscribed hexagon is composed of six equilateral triangles with side r.

    Its area is: . A_{hex} \;=\;6 \times \frac{\sqrt{3}}{4}\,r^2 \;=\;\frac{3\sqrt{3}}{2}\,r^2





    Code:
                  * o *
              *    /|\    *
            *     / | \     *
           *     /  |r \     *
                /   |   \
          *    /    |    \    *
          *   /  r  *  r  \   *
          *  /   *     *   \  *
            / *           * \
           o- - - - - - - - -o
            *               *
              *           *
                  * * *
    It can be shown that the side of this triangle is: . s = \sqrt{3}\,r

    Then its area is: . A_1 \;=\;\frac{\sqrt{3}}{4}\left(\sqrt{3}\,r\right)^2 \;=\;\frac{3\sqrt{3}}{4}r^2





    Code:
                            o
                           /|\
                          / | \
                         /  |  \
                        /   |   \
                       /    |    \
                      /   * * *   \
                     /*     |     *\
                    *       |       *
                   *        |     o  *
                  /         |   o     \
                 /*         | o  r    *\
                / *         *         * \
               /  *                   *  \
              /                           \
             /     *                 *     \
            /       *               *       \
           /          *           *          \
          o - - - - - - - * * *- - - - - - - - o
    It can be shown that the side of this triangle is: 2\sqrt{3}\,r

    Its area is: . A_2 \;=\;\frac{\sqrt{3}}{4}\,(2\sqrt{3}\,r)^2 \:=\:3\sqrt{3}\,r^2




    If A_{hex} is the mean proportional between A_1 and A_2

    . . then: . \left(A_{hex}\right)^2 \;=\;\left(A_1\right)\cdot\left(A_2\right)


    We have: . \left(\frac{3\sqrt{3}}{2}\,r^2\right)^2 \;=\;\left(\frac{3\sqrt{3}}{4}\,r^2\right)\cdot\le  ft(3\sqrt{3}\,r^2\right)

    . . which simplifies to: . \frac{27}{4}\,r^4 \;=\;\frac{27}{4}\,r^4 . . . it's true!

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