Find the area of the $AMRQ$ region
Solution 1:
Hint: $QFND$ is a parallelogram so altitude from $Q$ and $N$ to $DF$ is equal and given the common base $DR$, $S_{\triangle RQD} = S_{\triangle RND}$. Similarly, $BMFP$ is a parallelogram and $S_{\triangle BRM} = S_{\triangle BRP}$
Solution 2:
BPFM and FNQD are paralelograms $ML = LP, QT=TN\\ \therefore S_{RBM}=S_{RBP}\\ S_{RQF}=S_{RNF}\\ 20+A+C+2B =S+A+C+2B\\ \therefore S = 20$
