Results 1 to 2 of 2

Math Help - Ratio of inscribed and circumscribed triangles

  1. #1
    Newbie
    Joined
    Sep 2009
    Posts
    14

    Ratio of inscribed and circumscribed triangles

    An equilateral triangle is inscribed in a circle, and the circle is inscribed in another equilateral triangle. Find the ratio of the area of the smaller triangle to that of the larger triangle.

    Im getting 1:3 but the answer is 1:4.

    Thank You
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,915
    Thanks
    779
    Hello, spred!

    An equilateral triangle is inscribed in a circle,
    and the circle is inscribed in another equilateral triangle.
    Find the ratio of the area of the smaller triangle to that of the larger triangle.
    Code:
                          A
      D ∆ - - - - - - - * ∆ * - - - - - - - ∆ E
         \          *    / \    *          /
          \       *     /   \     *       /
           \     *     /     \     *     /
            \         /       \         / 
             \  *    /         \    *  /
              \ *   /     ∆ O   \   * /
               \*  /      |      \  */
                \ /       |       \ /
               C ∆  - - - * - - -  ∆ B
                  *       D       /
                   \*           */
                    \   * * *   /
                     \         /
                      \       /
                       \     /
                        \   /
                         \ /
                          ∆
                          F

    The center of the circle is O.
    The inscribed triangle is ABC.
    The circumscribed triangle is D{E}F.
    O is the centroid of \Delta ABC and \Delta D{E}F.

    Draw segment OB.

    In right triangle ODB,\,\text{ let }r = OD.
    . . Then the altitude AD of \Delta ABC is  3r.

    Since \angle ODB = 30^o\!:\;\;OB = 2r
    . . Then the altitude DB of \Delta D{E}F is 6r


    The ratio of the altitudes is: . 3r: 6r \:=\: 1: 2

    Therefore, the ratio of the area is: . 1^2:2^2 \;=\;1:4

    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 1
    Last Post: June 29th 2011, 11:26 PM
  2. Replies: 1
    Last Post: August 28th 2010, 03:48 AM
  3. Circumscribed/inscribed area problem
    Posted in the Calculus Forum
    Replies: 2
    Last Post: April 15th 2010, 12:18 PM
  4. Hyperbolic triangles and inscribed circles
    Posted in the Geometry Forum
    Replies: 0
    Last Post: April 8th 2010, 01:36 PM
  5. Replies: 3
    Last Post: July 21st 2008, 05:58 AM

Search Tags


/mathhelpforum @mathhelpforum