Disk's Circular Path

Let $D$ be the disk and $P$ be the circular path.

The disk had an area of $1$ squared unit. We can figure out the disk's circumference with this information.

$\begin{aligned} D_{A} &=1 \\ πr^{2} &=1 \\ r^{2} &= \frac{1}{π} \\ r &= \sqrt{\frac{1}{π}} \small \text{or} \frac{\sqrt{1}}{\sqrt{π}} \small \text{or} \frac{\sqrt{π}}{π} \end{aligned}$

With the radius given, we can solve for $D_{C}$ or the disk's circumference.

$\begin{aligned} D_{C} &= 2πr \\ &= 2π(\frac{\sqrt{π}}{π}) \\ &= 2\sqrt{π} \end{aligned}$

It was said that $P_{C}$ or the path's length is $π$ times $D_{C}$ .

$\begin{aligned} P_{C} &= D_{C}π \\ &= (2\sqrt{π})π \\ &= 2\sqrt{π^{3}} \end{aligned}$

We turn $P$ into a circle. If the path's length is $P_{C}$ or the circle's circumference, we can now get $P_{A}$ or the area of the circle.

$\begin{aligned} 2πr &= 2\sqrt{π^{3}} \\ r &= \frac{2\sqrt{π^{3}}}{2π} \\ r &= \sqrt{π} \end{aligned}$

$\begin{aligned} P_{A} &= πr^{2} \\ &= π(\sqrt{π})^{2} \\ &= π^{2} \end{aligned}$

So the area of the imaginary circle is $π^{2}$ which is also equal to $\frac{\sqrt{π^{5}}}{\sqrt{π}}$ .

$π^{2} \cdot \frac{\sqrt{π}}{\sqrt{π}} = \frac{π^{2}\cdot\sqrt{π}}{\sqrt{π}} = \frac{π^{2}\cdotπ^{\frac{1}{2}}}{π^{\frac{1}{2}}} = \frac{π^{\frac{5}{2}}}{π^{\frac{1}{2}}} = \boxed{\frac{\sqrt{π^{5}}}{\sqrt{π}}}$

Since $\frac{\sqrt{π}}{\sqrt{π}}$ is equals to $1$ , we don't actually change the value when it's multiplied to $π^{2}$ .