a) Rotate $S$ about $N, P, Q, M$ respectively to get $S_N, S_P, S_Q, S_M$. Note that $S_NS_PS_QS_M$ is a rectangle. Then consider a homothety with center $S$ and factor 1/2 that maps $S_N\mapsto N, S_P\mapsto P, S_Q\mapsto Q, S_M\mapsto M$. b) Let $DS=m, AS=n$. Then by Pythagorean Theorem on $ADS$ and $CBS$, $m^2+n^2=25$ and $(10-n)^2+(11-m)^2=100$. Solving, $(m,n)=(3,4)$. So, considering the lengths in $S_NS_PS_QS_M$, we have $SO=\sqrt{\left(\dfrac{11}{2}-3\right)^2+(5-4)^2}=\dfrac{\sqrt{29}}{2}$.

The quadrilateral formed by connecting the midpoints of another quadrilateral is always a parallelogram. Let a=MN=OP and let b=MP=ON. Since the diagonals of ABCD are perpendicular, the area of ABCD is half the product of the lengths of the diagonals, which is 2ab. However, the area of ABCD is also the sum of the areas of triangles AMN, BNO, CPO, and DPM, and quadrilateral MNOP. The sum of the areas of the triangles is ab, which means that the area of MNOP is also ab, which is the product of the lengths of its sides. Therefore, MNOP is a rectangle.