r o raio do menor círculo
R o raio do semi-círculo.\\
[tex3]\mathsf{
T (ponto ~de~ tangência) \implies T ponto ~médio~ do~ arco~ OPB.\\
OO1=r\sqrt2,R=OT=r(1+\sqrt2)\\
mas~ QO=r \therefore : r+PQ=R⇒r+\sqrt2=r+\sqrt2⇒r=1 ~e~ R=1+\sqrt2\\
Trace TH \perp PO. \\
TH=x\\
T.PIt: : 2x^2=(1+\sqrt2)^2⇒x=\frac{2+\sqrt2}{2} . \\
S_{\triangle PQT}={\sqrt 2×({2+\sqrt 2 \over 2})\over 2}={1+\sqrt 2 \over 2}\\
S△PQT=2–√×(2+2√2)2=1+2–√2S\\
S_{set(POT)}={{{\pi \over 4}×(1+\sqrt 2)^2}\over2}=\pi ×({3+2\sqrt 2 \over 8})\\
S_{seg(PT)}=S_{set(POT)}-S_{\triangle POT}=\pi ×({3+2\sqrt 2 \over 8})-{3\sqrt 2 +4 \over 4}\\
\therefore S_{sombreada}=S_{seg(PT)}+S_{\triangle PQT}=\pi ×({3+2\sqrt 2 \over 8})-{3\sqrt 2 +4 \over 4}+{1+\sqrt 2 \over 2}\\
=\boxed{\color{red}\pi ({3+2\sqrt 2 \over 8})-({2+\sqrt 2 \over 4})}
}[/tex3]
(Solução:NIgrumCibum -
viewtopic.php?t=89414)