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Math Help - square/circle

  1. #1
    Senior Member sfspitfire23's Avatar
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    square/circle

    Given that the area of a square and a circle are equal. Express circumference of circle a in terms of the perimeter of the square p.


    So, we know that
    \pi r^2=side^2. Now, s=\sqrt{\pi}r. Then 4s=4\sqrt{\pi}r. But 4s=perimeter. So, p=4\sqrt{\pi}r. Solve for r and we get r=\frac{p}{4\sqrt{\pi}}. Then plug this in to the equation of circumference to get the answer.

    Correct?
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  2. #2
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    Quote Originally Posted by sfspitfire23 View Post
    Given that the area of a square and a circle are equal. Express circumference of circle ‘a’ in terms of the perimeter of the square ‘p’.


    So, we know that
    \pi r^2=side^2. Now, s=\sqrt{\pi}r. Then 4s=4\sqrt{\pi}r. But 4s=perimeter. So, p=4\sqrt{\pi}r. Solve for r and we get r=\frac{p}{4\sqrt{\pi}}. Then plug this in to the equation of circumference to get the answer.

    Correct?
    Yes!
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  3. #3
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    Hello, sfspitfire23!


    Given that the area of a square and a circle are equal.
    Express circumference of circle C in terms of the perimeter of the square P.

    The areas are equal: . \pi r^2 \:=\:s^2 .[1]

    Since P \,=\,4s \quad\rightarrow\quad s \,=\,\frac{P}{4}

    Substitute into [1]: . \pi r^2 \:=\:\left(\dfrac{P}{4}\right)^2 \quad\Rightarrow\quad \pi r^2 \:=\:\dfrac{P^2}{16}

    . . r^2 \:=\:\dfrac{P^2}{16\pi} \quad\Rightarrow\quad r \:=\:\sqrt{\dfrac{P^2}{16\pi}} \quad\Rightarrow\quad r \:=\:\dfrac{P}{4\sqrt{\pi}}


    Multiply both sides by 2\pi\!:\;\;2\pi\cdot r \:=\:2\pi\cdot\dfrac{P}{4\sqrt{\pi}}


    Therefore: . C \:=\:\dfrac{P\sqrt{\pi}}{2}

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