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Math Help - Finding the radius of a circle inscribed in a sector.

  1. #1
    Member Furyan's Avatar
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    Finding the radius of a circle inscribed in a sector.

    Hello

    I have a question where I need to find the area of a circle inscribed, I think that's the right word, in a sector. The circle touches both straight edges of the sector and the arc. I know the angle of the sector, \dfrac{\pi}{3} and the lengths of the two straight edges, 6 cm. I figure I need to find the radius of the circle.

    I'm just after a hint. Thank you.
    Last edited by Furyan; April 13th 2012 at 02:36 PM.
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  2. #2
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    Re: Finding the radius of a circle inscribed in a sector.

    think properties of 30-60-90 triangles ...
    Attached Thumbnails Attached Thumbnails Finding the radius of a circle inscribed in a sector.-circlesector.jpg  
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  3. #3
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    Re: Finding the radius of a circle inscribed in a sector.

    Draw the bisector of the angle and connect the circle center with the points where the circle touches the angle's sides.Then use the fact that the resulting triangles are right-angled.
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    Re: Finding the radius of a circle inscribed in a sector.

    Hello, Furyan!

    I need to find the area of a circle inscribed in a sector.
    The circle touches both straight edges of the sector and the arc.
    Angle of the sector is \tfrac{\pi}{3} and the lengths of the two straight edges, 6 cm.
    I figure I need to find the radius of the circle. . Right!

    Code:
                              B
                            * o *
                    *   *     |     *   *
                 *    *       |       *    *
             A o     *        |r       *     o C
                \             |             /
                 \  *         |O        *  /
                  \ *         o         * /
                   \*         | *  r    */
                    \         |   *     / 
                     *        |     *  *
                   6  *       |       o
                       \*     |     */ Q
                        \   * * *   /
                         \    |    /
                          \   |   /
                           \30|30/
                            \ | /
                             \|/
                              o
                              P
    We have sector ABCP:\;\angle APC = 60^o,\;AP = BP = CP = 6.

    Inscribed circle: center O, radius OB = OQ = r.

    In right triangle OQP\!:\;\angle OPQ = 30^o,\;OQ = r.
    . . Hence: . OP = 2r


    Since BP = 6r, we have: . r + 2r \,=\,6

    Thanks from Furyan
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  5. #5
    Member Furyan's Avatar
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    Re: Finding the radius of a circle inscribed in a sector.

    Quote Originally Posted by skeeter View Post
    think properties of 30-60-90 triangles ...
    Thank you Skeeter. I hadn't spotted it was a 'special' triangle and the 2:1 connection. The graph was very helpful.

    Thank you.
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  6. #6
    Member Furyan's Avatar
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    Re: Finding the radius of a circle inscribed in a sector.

    Quote Originally Posted by emakarov View Post
    Draw the bisector of the angle and connect the circle center with the points where the circle touches the angle's sides.Then use the fact that the resulting triangles are right-angled.
    Thank you emakarov, I had drawn the angle bisector and was wrestling with those right triangles.

    Thank you.
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  7. #7
    Member Furyan's Avatar
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    Re: Finding the radius of a circle inscribed in a sector.

    Quote Originally Posted by Soroban View Post
    Hello, Furyan!


    Code:
                              B
                            * o *
                    *   *     |     *   *
                 *    *       |       *    *
             A o     *        |r       *     o C
                \             |             /
                 \  *         |O        *  /
                  \ *         o         * /
                   \*         | *  r    */
                    \         |   *     / 
                     *        |     *  *
                   6  *       |       o
                       \*     |     */ Q
                        \   * * *   /
                         \    |    /
                          \   |   /
                           \30|30/
                            \ | /
                             \|/
                              o
                              P
    We have sector ABCP:\;\angle APC = 60^o,\;AP = BP = CP = 6.

    Inscribed circle: center O, radius OB = OQ = r.

    In right triangle OQP\!:\;\angle OPQ = 30^o,\;OQ = r.
    . . Hence: . OP = 2r


    Since BP = 6r, we have: . r + 2r \,=\,6

    Thank you Soroban, I have to admit that I hadn't quite got there.

    BP = 3r, though. Right?

    Thank you.
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