Four equal circles intersect: What is the area of the small shaded portion and its height In terms of $R$ which is the radius of all four circles, what is the area of the intersection region of these four equal circles and the height of the marked arrow in the figure? The marked arrow is along the line CD, also the midpoint of all the circles are points A, B, C and D. Looking for a very short intuitive solution. I have checked similar questions on this site for example this and this. 

 A: 
Notice that the area of the equilateral triangle with edge $R$ plus $\frac12$ the area of the shaded region is $\frac16$ the area of the circle. The height of the triangle is $\frac{R\sqrt3}{2}$ and the area is $\frac12\cdot R\cdot\frac{R\sqrt3}{2}=\frac{R^2\sqrt3}{4}$. The area of the shaded region is:
$$2\left(\frac{\pi R^2}{6}-\frac{R^2\sqrt3}{4}\right)=\frac{2\pi R^2}{6}-\frac{3R^2\sqrt3}{6}=\frac{R^2(2\pi-3\sqrt3)}{6}$$
Also notice that the height of the triangle plus $\frac12$ the height of the shaded area is equal to the radius of the circle. The height of the shaded area is:
$$2\left(R-\frac{R\sqrt3}{2}\right)=2R-R\sqrt3=R(2-\sqrt3)$$
A: 
If you repeat this pattern infinitely, you'll find that each circle consists of $12$ of these (American!) football shaped areas, together with $6$ 'triangular' areas in between (see red circle in figure). So, setting the area of the footballs to $A$, and that of the triangles to $B$, we have:
$\pi=12A +6B$  (I am setting radius to $1$ ... you can easily adjust for $R$)
OK, now consider the green rectangle formed by four of the points on a circle. We see that the height of such a rectangle is equal to the radius, so that is $1$, and the width is easily found to be $\sqrt{3}$, and this area includes $4$ whole footballs, $4$ half footballs, $2$ whole 'triangles', and $4$ half triangles. So:
$\sqrt{3}=6A+4B$
Now you can easily solve for $A$
A: $AC=BC=AB=AD=BD=R$.Now $AB$ and $CD$ are perpendicular bisectors of each other, say at $O$.Then $DO^2=AD^2- AO^2= R^2- ({R \over 2})^2$ i.e. $DO={ \sqrt{3}R \over 2}$.Then $height=2(R- { \sqrt{3}R \over 2})=(2- \sqrt{3})R$.
Now if the shaded area be $∆$, as $\angle ADB=60°$ , ${∆ \over 2}={πR^2 \over 6}- { \sqrt{3}R^2 \over 4}$(it's area of $∆ABD$)so, required shaded region area $∆={2π-3√3 \over 6}R$.
