How do I find the area of this region? A square with edge length 2 cm has semicircles drawn on each side.
Find the total area of the shaded region. 
Here is an image of the diagram shown :
 
Please show your work in pictures, numbers, words, anything. (Try to keep it to a Grade 8 level too)
 A: One sees four semi-circles. Each delineates a semi-disk. Summing the areas of these four semi-disks, one counts twice each purple region and once each white region. 
Hence the purple area plus the area of the square with side $2$ is twice the area of a disk with diameter $2$, that is, $\color{purple}{\mathbf{purple\ area}}+2^2=2\cdot\frac14\pi\cdot2^2$.
Finally, $\color{purple}{\mathbf{purple\ area}}=2\pi-4$.
A: There are $8$ circular segments, each of area
$$\frac{\pi}{2} r^2 (\theta - \sin{\theta}) $$
where $r$ is the radius, and $\theta$ is the angle subtended at the center by the segment.  Here, from the geometry, $r$ is clearly $1 \, \mathrm{cm}$ and $\theta = \frac{\pi}{2}$, each segment spanning half a semicircle.
Therefore the shaded area is
$$8 \frac{\pi}{2} (1 \, \mathrm{cm}^2) \left ( \frac{\pi}{2} - 1 \right ) = (2 \pi - 4) \, \mathrm{cm}^2$$
A: The square has area $4$. 
The top and bottom white region together have area $4$ minus $2$ times the area of the semi circle, that is, $4-2\times \pi r^2/2=4-\pi$ (since $r=1$). 

So, dividing by $2$, the top blue region has area $2-\pi/2$. 
If you remove this area from the area of the top semi circle you obtain the area of the two top red parts. 

Double it to obtain the red area.
A: Move one half of a leaf to the circle:
$\hspace{3cm}$
The area of the quarter circle is $\frac\pi4$; the area of the triangle is $\frac12$. Thus, the area of the half-leaf (in red) is $\frac\pi4-\frac12$. There are two per leaf, therefore the area of four leaves is
$$
8\left(\frac\pi4-\frac12\right)=2\pi-4
$$
A: A layman's (mine) approach to solving it:
There are 4 half-circles contained in the square. The 4 half-circles have the same area as two whole circles with the same radius. The radius (half the length of the diameter, which is the length of a side of the cube) is 1 and the area of a circle is $\pi r^2$, so the area of the four half-circles is $2 ( \pi * (1)^2)$, which is $2\pi$. The square can only hold an area of $4$ and all its space is used, so the excess area of the circles' area must be contained in overlap. The the area not able to be contained be the square without overlap is given by $2\pi - 4$.
This, of course, only works out nicely because ALL the area inside the square is being used. Were part of the square unused, you would have to add the area of the unused portion into the calculation of the overlap.
A: 
Try this simple approach instead.
From Fig. A area of the shaded region is $A_{A}(4-\pi)$
Similarly with Fig B the area is $A_{B} = (4-\pi)$
Now when you subtract it from the whole you get the four shaded areas in the given question. The shaded area is $A = 4 -(4-\pi)-(4-\pi)= 2\pi-4$
