TITLE: find area of dark part
QUESTION [8 upvotes]: 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

REPLY [3 votes]: 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:
$$
S = \dfrac{1}{8} (S_{\mathrm{large}} - 4\cdot\dfrac{3}{4}S_{\mathrm{small}} - S_{\Box}) = 
\dfrac{1}{8}(\pi R^2 - 3\pi r^2 - 4r^2) = \dfrac{\pi\sqrt{2}-2}{4}r^2.
$$