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Math Help - concentric circles

  1. #1
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    concentric circles

    An archery target consists of three concentric circles. The outer rim of the target has a radius of 1.2m. Find the radii of the two inner circles so that the three areas( central circle and two rings are equal)
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  2. #2
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    total area = \pi(1.2)^2

    inner circle ... \pi r_1^2 = \frac{\pi(1.2)^2}{3}

    middle ring ... \pi(r_2^2 - r_1^2) = \frac{\pi(1.2)^2}{3}

    outer ring ... \pi[(1.2)^2 - r_2^2] = \frac{\pi(1.2)^2}{3}
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  3. #3
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    supposely the answer is 0.4m,0.8m,1.2m

    but, that cant be right, eh
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  4. #4
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    Hello, requal!

    An archery target consists of three concentric circles.
    The outer rim of the target has a radius of 1.2m.
    Find the radii of the two inner circles so that the three areas
    (central circle and two rings) are equal)

    Let a = smallest radius.
    Let b = next larger radius.
    We know that c = 1.2 is the largest radius.

    The area of the central circle is: . \pi a^2 .[1]

    The area of the middle ring is: . \pi b^2 - \pi a^2 \:=\:\pi(b^2-a^2) .[2]

    The area of the outer ring is: . \pi(1.2^2) - \pi b^2 \:=\:(1.44 - b^2)\pi .[3]


    Since the areas are equal, [1] = [2] gives us:

    . . \pi a^2 \:=\:\pi(b^2-a^2) \quad\Rightarrow\quad b^2 - 2a^2 \:=\:0 .[4]


    And [2] = [3] gives us:

    . . \pi(b^2-a^2) \:=\:\pi(1.44 - b^2) \quad\Rightarrow\quad 2b^2 - a^2 \:=\:1.44 .[5]


    Solve the system: . \begin{array}{cccc}b^2-2a^2 &=& 0 & {\color{blue}[4]} \\ 2b^2 - a^2 &=& 1.44 & {\color{blue}[5]} \end{array}

    . . and we get: . \boxed{a \:=\:\frac{2\sqrt{3}}{5}} \quad\boxed{ b \:=\:\frac{2\sqrt{6}}{5}}

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  5. #5
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    so, the teacher was wrong after all,
    thanks
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