Find the Area of the lens of 2 overlapping circles I'm trying to find the area of a lens of 2 overlapping circles of the same size.
The circles are both $4$ feet diameter (Radius $2$ feet) and the distance between both radii is $2.75$ feet.  After using the formula posted for $2$ circles of the same radii (found here) using values of radius $a=2$, offset $d = 2.75$ and a bonus height $h=3$. 
After running all my calculations I got the value $A=7.1$ feet. However looking at the total area of the circle $4$ feet across being $12.57$ feet, then $7.1$ feet is the greater portion of the circle not the lens. 
My question: Is the value calculated the in the equation of the Area of the remaining circle (lens would be the difference) or is $7.1$ in fact the Area of the lens? I'm having trouble confirming my answer is correct. My goal is to find the total area of the highlighted area shown in the photo minus the overlap areas in square feet. These are tables in a store, and need the total square feet to floor space these tables took up. So over lap value is important when subtracting the area of the table alone. ALL OVERLAPPING AREAS ARE EQUAL, ALL TABLES ARE EQUAL. 
Thank you.

So you say that the two radii are equal. Then $R = r = 2$ feet. The offset is $d = 2.75$ feet. Thus, if I apply the same formula, I get
\begin{align*}
A &= 2^2\cos^{-1}\left(\frac{2.75^2+2^2-2^2}{2(2.75)(2)}\right)+2^2\cos^{-1}\left(\frac{2.75^2+2^2-2^2}{2(2.75)(2)}\right)\\
&\qquad-\frac{1}{2}\sqrt{(-2.75+2+2)(2.75+2-2)(2.75-2+2)(2.75+2+2)}\\
&= 8\cos^{-1}\left(\frac{2.75}{4}\right)-\frac{1}{2}(2.75)\sqrt{(4-2.75)(4+2.75)}\\
&= 2.50803.
\end{align*}
So, assuming this is the right formual, it looks like the overlap is about $2.5\text{ ft}^2$.
