Find the area of ​the $AMRQ$ region

Question

For reference: In figure : $ABCD$, it is a parallelogram $MN \parallel AD, PQ \parallel AB$; if: the area of $​​RPCN$ is $20m^2$ calculate the area of ​​the $AMRQ$ region (Answer:$20m^2$)

My progress:

$BM=a\\ AM=b\\ BP=m\\ PC=n\\ S_{MRA}=A,S_{ARQ}=B,S_{PRC}=M, S_{CRN}=N\\ A+B=x\\ M+N=20\\ S_{BMR} = \frac{Aa}{b}\\ S_{DRQ}=\frac{Bn}{m}\\ S_{NDR}=\frac{Nb}{a}\\ S_{BPR}=\frac{Mm}{n}\\ S_{ABD}=S_{BCD}\\ x+\frac{Aa}{b}+\frac{Bn}{m}=20+\frac{Mm}{n}+\frac{Nb}{a}$

...???

Answer 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 $S_{\triangle BRM} = S_{\triangle BRP}$

Answer 2

Transform the parallelogram into a unit square, then the ratio of areas is preserved.

The images of the various points will be as follows

$A = (0, 0)$

$B = (0, 1)$

$C = (1,1)$

$D = (1, 0)$

$R = (s, 1- s)$

$Q = (t, 0)$

$P = (t, 1)$

$N = (1, r)$

$M = (0, r)$

$F = (t, r)$

Since point $F$ lies on the diagonal, then $r = 1- t$

Now apply the so-called shoelace formula, to both quadrilaterals

$[PRCN] = (1, r) \times (1,1) + (1, 1) \times (t, 1) + (t, 1) \times (s, 1- s) + (s, 1- s) \times (1, r) $

where $(a, b ) \times (c, d) = a d - b c $

This reduces to

$[PRNC] = 1- r - ts + s r$

Similarly calculation for [AMRQ] results in

$[AMRQ] = (t, 0) \times (s, 1- s) + (s, 1 - s) \times (0, r) = t - t s + s r $

But as was stated above $r = 1 - t$ , so $t = 1 - r$ and therefore the two areas are equal.

Answer 3

BPFM and FNQD are paralelograms $ML = LP, QT=TN\\ \therefore S_{RBM}=S_{RBP}\\ S_{RQF}=S_{RNF}\\ 20+A+C+2B =S+A+C+2B\\ \therefore S = 20$