trigonometry
Please help calculate the total un-overlapped area of union inside a boundary consisting of perimeters of 3 different circles as shown. The two larger circles are 4.25 mm in diameter and the center circle is 3.28 mm.
All the circles are centered on x-axis, the distance between the 2 largest circle centers is 1.34 mm. The middle smaller circle is centered along y-axis.
Here's a picture of the case $n=5$. 
If $R$ is the radius of the big circle and $r$ the radii of the small circles, do you see that the isosceles triangle has two sides $R-r$ and one side $2r$?
From the lower left-hand corner of the rectangle (let's call that point $A$), consider the diagonal line as a secant line of the left-hand circle, intersecting the circle at points $B$ and $C$, where $B$ is between $A$ and $C$. Also consider one of the edges of the rectangle adjacent to $A$ as a tangent line touching the circle at $D$.
Then by a theorem about tangent and secant lines from a point outside a circle, we have the following relationship of the lengths of the segments from $A$ to each of the three points $B$, $C$, and $D$: $$ (AD)^2 = AB \times AC. \tag1 $$
It is easy to find that $AD = \frac12a$. Now let $E$ be the midpoint of the bottom side of the rectangle; then $AC$ is the hypotenuse of right triangle $\triangle AEC$, which has legs $a$ and $\frac12a$, and therefore $AC = (\frac12\sqrt5)a$.
We can then use Equation $(1)$ to find the length $AB$, so we can find the length of the chord $BC$; from that chord and the radius of the circle we can get the angle of $S_1$ at the center of the circle.
Best Answer
Let's start by finding the combined area of the two larger circles. The area of one of the larger circles, which I will call $A_O$, is:
$$ \begin{align*} A_O &= \frac14\pi d^2\\ &= \frac14\pi(4.25)^2\\ &\approx14.186 \end{align*} $$
Thus the area of both circles, ignoring their intersection for a moment, is $2A_O$. To find the area between them, we can use the formula for the area of a symmetric lens,
$$A_\text{lens} = R^2(\theta-\sin\theta), \tag{*}\label{*}$$
where $R$ is the radius of the circles forming the lens, and $\theta$ the central angle in radians. We know $R = 4.25$, but we need to find $\theta$. To do so, draw in a few lines on the diagram:
(Note that I left out the smaller circle for simplicity's sake.) In the diagram, $\theta = m\angle AOB = 2\alpha$, and $R = OA = OC = OB$. We can find $\theta$ through $\triangle AOD$. The length of line segment $OD$ is equal to $OC-CD$, or the radius minus half the distance between the two circles. We also know the length of line segment $OA$, since it is the radius, so we can write:
$$ \begin{align*} \alpha &= \arccos\left(\frac{OD}{OA}\right)\\ &= \arccos\left( \frac{2.125-\frac12(1.34)}{2.125} \right)\\ &\approx .816596 \ \text{radians} \end{align*} $$
Since $\theta = 2\alpha$, $\theta=1.63319$. We can now use $\eqref{*}$:
$$ \begin{align*} A_\text{lens} &= R^2(\theta-\sin\theta) \\ &= (2.125)^2(1.63319-\sin1.63319)\\ &\approx 2.868 \end{align*} $$
Thus, the area between the two larger circles, $A_L$, is:
$$ \begin{align*} A_L &= 2A_O-A_\text{lens}\\ &= 2(14.186)-2.868\\ &= 25.504 \end{align*} $$
Now we have to deal with the two small regions created by the smaller circle. For these, I will be using the (rather lengthy) formula for the area of an asymmetric lens,
$$ \begin{align*} A_\text{asy} = & \ r^2\arccos\left( \frac{d^2+r^2-R^2}{2dr} \right) + R^2\arccos\left( \frac{d^2+R^2-r^2}{2dR} \right) - \\ & \frac12 \sqrt{(-d+r+R)(d+r-R)(d-r+R)(d+r+R)}, \tag{**}\label{**} \end{align*} $$
where $r$ is the radius of the smaller circle, $R$ is the radius of the larger circle, and $d$ is the distance between the centers of the two circles. Here, $r=1.64$, $R = 2.125$, and $d=1.455$. Observe the diagram below.
The light blue region is the asymmetric lens we will be finding the area of, and we are looking for the area of the two green regions. Using \eqref{**}, we find the area of one asymmetric lens to be $A_\text{asy}\approx5.650$, and thus the area of two asymmetric lenses is $2A_\text{asy}=11.300$.
Notice that the intersection of the two asymmetric lenses in the diagram is the symmetric lens from before. Thus, the area of the light blue and dark blue region is:
$$ \begin{align*} A_\text{blue} &= 2A_\text{asy}-A_\text{lens}\\ &=11.300-2.868\\ &=8.432 \end{align*} $$
Now, the area of the green regions is equal to the area of the small circle minus the area of the blue regions. The area of the small circle is:
$$ \begin{align*} A_o &= \pi r^2 \\ &= \pi(1.64)^2\\ &\approx 8.450 \end{align*} $$
And so the area of the green regions is:
$$ \begin{align*} A_\text{green} &= A_o-A_\text{blue}\\ &=8.450-8.432\\ &=0.018 \end{align*} $$
Finally, the area of the entire shape is the area of the two large circles plus the area of the green regions, which is:
$$ \begin{align*} A &= A_L + A_\text{green} \\ &= 25.504 + 0.018\\ &= \boxed{25.522 \text{ mm}^2} \end{align*} $$