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Math Help - please please help! i have finals tomorrow!

  1. #1
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    please please help! i have finals tomorrow!

    well my question is how to calculate the area of the circle when its in a triangle and i only have the units for the triangle and not the circle.
    here is the actual problem: all the sides of the triangle is 8 units each. then there is a circle perfectly placed inside. and i need to find the area of the circle.

    so far i think i need to find the area of a triangle first, which i did and i just need to get going...
    Last edited by svetka; January 28th 2009 at 08:04 PM. Reason: thought of more things how to solve my program
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  2. #2
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    Quote Originally Posted by svetka View Post
    well my question is how to calculate the area of the circle when its in a triangle and i only have the units for the triangle and not the circle.
    here is the actual problem: all the sides of the triangle is 8 units each. then there is a circle perfectly placed inside. and i need to find the area of the circle.

    so far i think i need to find the area of a triangle first, which i did and i just need to get going...
    In an equilateral triangle of sidelength a the length of the median is \frac{\sqrt{3}}{2} \cdot a

    The radius of the circle is 1/3 times the length of the median because the centre of the circle is the centroid of the triangle and hence divides the median in the ratio 2:1.

    Therefore the radius of the circle is \frac{a \sqrt{3}}{6}.

    In your question a = 8.
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  3. #3
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    Hello, svetka!

    Another approach . . .


    A circle is inscribed in an equilateral triangle of side 8.
    Find the area of the circle.
    Code:
                            *
                           / \
                          /   \
                         /     \
                        /       \
                       /         \
                      /   * * *   \
                     /*           *\
                    *               *
                   *  \           /  *
                  /    r\       /r    \
                 /*       \   /       *\
                / *         *         * \
               /  *         |         *  \
              /             |             \
             /     *        |r       *     \
            /       *       |       *       \
           /          *     |     *          \
          *-------------*-*-*-----------------*
          : - - - - - - - - 8 - - - - - - - - :

    The area of an equilateral triangle of side x is: . A \:=\:\frac{\sqrt{3}}{4}x^2
    We have x = 8, so: . A \:=\:\frac{\sqrt{3}}{4}(8^2) \:=\:16\sqrt{3}


    Formula: .The area of a triangle is: . A \:=\:\tfrac{1}{2}pr
    . . . . . . . where p is the perimeter, and r is the radius of the inscribed circle.

    The perimeter of this triangle is 24, so we have: . \tfrac{1}{2}(24)r \:=\:16\sqrt{3}

    Hence: . 12r \:=\:16\sqrt{3} \quad\Rightarrow\quad r \:=\:\frac{4\sqrt{3}}{3}


    Now you can find the area of the circle . . .

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