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Math Help - Equilateral Triange inside a circle

  1. #1
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    Equilateral Triange inside a circle

    Hello all,

    I'm hoping someone can help with this one, I'm not sure why I'm not seeing it . I've attached a picture from my text book and an answer I've found but don't quite understand.

    Question:
    An equilateral triangle where each side's length equals x is inscribed in a circle of radius r. Show that:

    <br />
r^2=\frac{x^2}{3}<br />


    An answer I've found already but don't understand:


    . Thus:

    . Solve for rē and you'll get:



    However, I'm not sure how one is supposed to know that r=\frac{2}{3}\cdot h everything else made sense though...

    Thanks,
    Skilo
    Attached Thumbnails Attached Thumbnails Equilateral Triange inside a circle-triangle.jpg  
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  2. #2
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    Quote Originally Posted by Skilo View Post
    Hello all,

    I'm hoping someone can help with this one, I'm not sure why I'm not seeing it . I've attached a picture from my text book and an answer I've found but don't quite understand.

    Question:
    An equilateral triangle where each side's length equals x is inscribed in a circle of radius r. Show that:

    <br />
r^2=\frac{x^2}{3}<br />


    An answer I've found already but don't understand:


    . Thus:

    . Solve for rē and you'll get:



    However, I'm not sure how one is supposed to know that r=\frac{2}{3}\cdot h everything else made sense though...

    Thanks,
    Skilo
    The area of the triangle can found using Heron's Formula:

    s = \frac{x + x + x}{2} = \frac{3x}{2}

    A = \sqrt{s(s - x)(s - x)(s - x)}

     = \sqrt{\frac{3x}{2}\left(\frac{3x}{2} - x\right)\left(\frac{3x}{2} - x\right)\left(\frac{3x}{2} - x\right)}

     = \sqrt{\frac{3x}{2}\left(\frac{x}{2}\right)\left(\f  rac{x}{2}\right)\left(\frac{x}{2}\right)}

     = \sqrt{\frac{3x^4}{16}}

     = \frac{\sqrt{3}x^2}{4}.


    We can also work out the area using the fact that the equilateral triangle is made up of three equal triangles of side lengths r, r, x, with an angle of 120^{\circ} between the sides of lengths r, r.

    So the area of the equilateral triangle will be

    A = 3\cdot \frac{1}{2}\cdot r \cdot r \cdot \sin{120^\circ}

     = \frac{3r^2}{2}\cdot \frac{\sqrt{3}}{2}

     = \frac{3\sqrt{3}r^2}{4}.


    Since the areas are equal, we have

    \frac{\sqrt{3}x^2}{4} = \frac{3\sqrt{3}r^2}{4}

    Therefore x^2 = 3r^2

    or r^2 = \frac{x^2}{3}.
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  3. #3
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    for that h = (3/2) r :

    First of all, the line from the center to the points of the equil triangle has 2 properties:
    a. the length is equal to radius of circle (everybody knows)
    b. the line bisect the angle of the equil triangle

    So, consider h is the height of the equil triangle
    h = r + s , assuming s is just the little portion extending from the line r.

    do you see sin 30 = s / r ?
    sin 30 = s / r
    (1/2) = s / r
    s = (1/2) r
    thus h = r + (1/2) r
    h = (3/2) r
    and r = (2/3) h

    To prove the properties above, it involves showing 2 trangles are SSS. Just try to join lines from the center and you will see how.
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  4. #4
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    Hello, Skilo!

    However, I'm not sure how one is supposed to know that r\,=\,\tfrac{2}{3}h
    Code:
                    A
       -          * o *
       :      *    /|\    *
       :    *     / | \     *
       :   *     /  |r \     *
       h        /   |   \
       :  *    /    |    \    *
       :  *   /     oG    \   *
       :  *  /   *  |  *   \  *
       :    / *     |     * \
       - B o- - - - o - - - -o C
            *       M       *
              *           *
                  * * *

    We have an equilateral triangle inscribed in a circle.

    The center of the circle G is the centroid,
    . . center of mass of the triangle,
    . . which divides the median AM in the ratio 2:1

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  5. #5
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    Equilateral triangle inside a circle

    Posted by Skilo

    Using Soroban's nice diagram apply the 30-60-90 triangle rule.
    If hypothenuse is 2 the side opposite the 30 angle =1. the side opposite the 60 angle = radical 3 What is angle ACG .Can you prove it.

    bjh
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  6. #6
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    Drop a perpendicular form the center to one of the left or right sides, and form a 30, 60, 90 triangle whose sides are in the ratio 1:2:sqrt(3) Then you have...

    r/(x/2) = 2/sqrt(3), and the result follows directly.
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  7. #7
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    Thanks everyone for all the help !
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  8. #8
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    Using the Soroban's diagram (thanks Soroban)

    The proves for:
    1. Why angles ABG, BAG, BCG, CBG, ACG, CAG are all 30 degree:
    - triangle ABC is a equil trangle, so AB=BC=AC
    - Angles ABC=CBA=CAB=60 degrees
    - AG=BG=CG=radius
    - So, triangles ABG, BCG, CAG are SSS
    - thus angles ABG=BAG=BCG=CBG=ACG=CAG=30 degree
    - so, BG bisect angle ABC, CG bisect angle BCA, AG bisect angle CAB

    2. After extending the line AG to M becoming AM, why AM perp to BC:
    conside triangle BAM and CAM
    - AB=AC
    - AM=AM
    - angle BAM = angle CAM
    so, triangle BAM and triangle CAM are SAS
    thus angle AMB = angle AMC = 90 degree
    so, AM perp to BC.

    So, AM is the height of the triangle ABC, passing the centre of the circle.
    let h = AM = AG + GM
    let r = AG = BG = radius
    let s = GM
    h = r + s
    sin 30 = GM / BG
    1 / 2 = s / r
    s = (1/2) r
    so h = r + s
    = r + (1/2) r
    = (3/2) r
    and r = (2/3) h
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