1. 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?

2. Originally Posted by sfspitfire23
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!

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