How to find the perimeter of a shaded region when one side isn't given?

Question

The problem is as follows:

Use the figure from below to find the orange shaded regions.

$\begin{array}{ll} 1.&16(4+\pi)\,\textrm{in}\\ 2.&8(12+\pi)\,\textrm{in}\\ 3.&4(16+\pi)\,\textrm{in}\\ 4.&12(8+\pi)\,\textrm{in}\\ \end{array}$

I'm confused exactly how to approach this problem?. The thing is how to get that side for the little square which is on the top left?.

Because the side has some displacement or off set from the top makes it difficult to get the radius.

I've attempted to do the following:

Assuming $a$ is the size of one side on the little square:

Where $P$ is perimeter:

$P=(8+4\sqrt{2}-a)\sqrt{2}\cdot \pi+4(8+4\sqrt{2}-a)+2(4)(a)$

But this expression is not possible to simplify. I mean the term $a$ doesn't cancel.

What can it be done here?. Can someone help me please?. Is there any trick here or what?.

Answer 1

Note that the diagonal of the inner square is the diameter of the circle. With $r$ being the radius of the circle and $i$ the side length of the inner square, you get according to Pythagoras

$$2i^2 =(2r)^2$$

Hence, you get the following system of equations for $a$ and $r$:

\begin{eqnarray*} \sqrt 2 r + a & = & 8+4 \sqrt 2\\ 2 r - a & = & 8+4 \sqrt 2\\ \end{eqnarray*}

Solving this gives $r=8$ and $a=8-4\sqrt 2$. So, the perimeter $P$ of the shaded region is $P = 16(\pi + 4)$.

Answer 2

Hint: if $r$ is the radius, the side of the large square is $r\sqrt 2$ and the side of the small square is $r-\frac{r\sqrt 2}{2}$.
Thus, $r+\frac{r\sqrt 2}{2}=8+4\sqrt 2$, $r=8$