I can't believe that it's a TST problem are you sure? It's really easy

sabbasizadeh wrote: I can't believe that it's a TST problem are you sure? It's really easy Of course I am sure!

Let the centre be $O$ and the diagonal intersection be $P$. Note that $K_a$ is tangent to $K_b$ at $P$. Then $\angle BDC=45$ so the area of the sector $BOC$ is $\frac14 \pi OB^2$. Let $E,F$ be midpoints of $AB,CD$, then clearly $E,P,O,F$ are collinear, and $FO=PE=\frac{a}{2}$ by simple length chase. The area of sector $EBP$ is $\frac14 \pi EB^2$ and the area of sector $CFP$ is $\frac14 \pi FC^2$. Thus their sum is $\frac14 \pi (EB^2+FC^2)=\frac14 \pi OB^2$ which is the area of sector $BOC$. Thus the required area is now just $S_{OAB}+S_{OCD}=2(\frac{a}{2})(\frac{c}{2})=\frac{ac}{2}$. The arc length of $BC$ is $\frac12 \pi OB$ and the arc lengths of $BP$ and $CP$ are $\frac12 \pi EP$ and $\frac12 \pi FP$. So now the required perimeter is $\pi(\frac{a}{2}+\frac{c}{2}+\frac{\sqrt{a^2+b^2}}{2})$.

Note that the trapezoid is isosceles. Let $O$ be the center of its circumcircle, $M$ and $N$ the midpoints of $AB$ and $CD$, $I$ the intersection of the diagonals; $K$, $K_a$, $K_b$ have centers $O$, $M$, $N$. Note that $\bigtriangleup ABP$ and $\bigtriangleup CDP$ are half-squares; $OB=OC$ and $\widehat {BOM}=\frac 1 2 \widehat{BOA}=\widehat{BDA}=\widehat {BCA}=\widehat{BCP}=\frac \pi 2 -\widehat {CBP}=\frac \pi 2-\widehat{CBD}=\frac \pi 2-\frac 1 2 \widehat{COD}=\frac \pi 2-\widehat{CON}=\widehat{NCO}$, so that $\bigtriangleup OBM\cong \bigtriangleup NCO$; therefore $\widehat{BOC}=\frac \pi 4$. The circumradius of $K$ is elementarily calculated, using the triangle $\bigtriangleup ABD$ with sides $a, \frac {a+c} {\sqrt 2}, \frac {\sqrt {a^2+c^2}} {\sqrt 2}$, to be $\frac {\sqrt {a^2+c^2}} 2$. Then we calculate the area we want as $2\left [ S(\stackrel{\frown}{BOC})+S(\bigtriangleup BOM)+S(\bigtriangleup NCO)-S(\stackrel{\frown}{CNP})-S(\stackrel{\frown}{BMP})\right ] =2\left [ \frac \pi 4 \frac {a^2+c^2} 4+2 \frac 1 2 \frac a 2 \frac c 2-\frac \pi 4 \frac {c^2} 4 -\frac \pi 4 \frac {a^2} 4 \right ] =\frac {ac} 2$. For the perimeter, in a similar fashion we get a value of $2\left [ \ell(\stackrel{\frown}{BOC})+\ell(\stackrel{\frown}{CNP})+\ell(\stackrel{\frown}{BOP})\right ] =2\left [ 2\frac \pi 4 \frac {\sqrt{a^2+c^2}} 2+2\frac \pi 4 \frac a 2+2 \frac \pi 4 \frac c 2 \right ] =\frac \pi 2 (\sqrt{a^2+c^2}+a+c)$.