find area of dark part let us consider following picture

we have following    informations.we have circular sector,central angle is  $90$,and in this sector there is inscribed small circle ,which touches arcs of sectors and radius,radius of this small circle is equal to  $\sqrt{2}$,we should find  area of  dark part.
my approaches is following,let us  connect radius  $\sqrt{2}$ to intersection points of small circle with radius of  big sector, we will get  square with length  $\sqrt{2}$,clearly area of dark part is  area of sector -area of square-area of small sector(inside small circle)and minus also area of small part below,which i think represent also sector  with central angle  $90$,but my question is  what is  radius of big sector?is it $2*\sqrt{2}$?or does  radius of small circle divide given  radius of big  sector into two    part?also please  give me hint if my approaches is  wrong or  good.thanks in advance
 A: Hints:


*

*Let $r = \sqrt{2}$ be the small radius.

*Big radius is equal to diagonal of small $r^2$ square plus $r$, that is $\sqrt{2}\cdot \sqrt{2} + \sqrt{2} = 2+ \sqrt{2}$.

*The small area near the origin has area of a square minus area of a circle divided by four.

*The dark area is whatever is left divided by two (the figure is symmetric).


I hope this helps ;-)
Edit: For check: $\frac{1}{2}\Bigg(\frac{\pi R^2}{4}-\pi r^2-\Big(r^2-\frac{\pi r^2}{4}\Big)\Bigg)$, where $R = (\sqrt{2}+1)r$.
A: 
Skeleton:
Denote small radius as $r$, large radius as $R$.
Then (looking at the image, diagonal) $2R = r+\sqrt{2}\cdot 2r + r = 2r+2\sqrt{2}r$, $\:$ so $$R=(1+\sqrt{2})r.$$
Square of black figure:
$$
S = \dfrac{1}{8} (S_{\mathrm{large}} - 4\cdot\dfrac{3}{4}S_{\mathrm{small}} - S_{\Box}) = 
\dfrac{1}{8}(\pi R^2 - 3\pi r^2 - 4r^2) = \dfrac{\pi\sqrt{2}-2}{4}r^2.
$$
A: Let $Y$ be the upper left corner of the picture, and let $O$ be the lower left corner, the centre of the quarter-circle.
Let $C$ be the centre of the small circle. Let the line $OC$ meet the big circle at $M$.
Draw a perpendicular from $C$ to $OY$, meeting $OY$ at $P$,
Note that by the Pythagorean Theorem, we have $OC=2$. Thus $OM=2+\sqrt{2}$. This is the radius of the big quarter-circle.
Now we know almost everything. Our target region is one-half of our quarter circle, minus triangle $OCP$, minus the little circle's circular sector $CMP$. 
The area of half our quarter-circle is easily found. So is the area of $\triangle OCP$. 
Finally, $\angle PCM$ is $135^\circ$, so the circular sector $CMP$ has area three-eighths of the area of the small circle.
