11. A paper rectangle $A B C D(A B=3, B C=9)$ was folded so that vertices $A$ and $C$ coincided. What is the area of the resulting pentagon?
$\square$ Answer: 19.5.

Fig. 4. Note that we folded along the perpendicular bisector $\ell$ of the segment $A C$. Let $M$ be the intersection point of the lines $\ell$ and $B C$, $N$ be the midpoint of $A C$, and the point $D$ became the point $D^{\prime}$. It is easy to notice that $B M=D^{\prime} K=4, A M=M C=A K=5, A N=3 \sqrt{10} / 2$ and $M N=N K=\sqrt{10} / 2$. Then $S\left(A B M K D^{\prime}\right)=3 \cdot 4+\sqrt{10} / 2 \cdot 3 \sqrt{10} / 2=19.5$.

(Additional problems Kangaroo 2014)