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Thread: 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
    $\displaystyle \pi r^2=side^2$. Now, $\displaystyle s=\sqrt{\pi}r$. Then $\displaystyle 4s=4\sqrt{\pi}r$. But $\displaystyle 4s=perimeter$. So, $\displaystyle p=4\sqrt{\pi}r$. Solve for $\displaystyle r$ and we get $\displaystyle r=\frac{p}{4\sqrt{\pi}}$. 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
    $\displaystyle \pi r^2=side^2$. Now, $\displaystyle s=\sqrt{\pi}r$. Then $\displaystyle 4s=4\sqrt{\pi}r$. But $\displaystyle 4s=perimeter$. So, $\displaystyle p=4\sqrt{\pi}r$. Solve for $\displaystyle r$ and we get $\displaystyle r=\frac{p}{4\sqrt{\pi}}$. 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 $\displaystyle C$ in terms of the perimeter of the square $\displaystyle P.$

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

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

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

    . . $\displaystyle 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 $\displaystyle 2\pi\!:\;\;2\pi\cdot r \:=\:2\pi\cdot\dfrac{P}{4\sqrt{\pi}}$


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

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