Find the area $x$ of the quadrilateral below In the figure, $ABCD$ is a square. If MN is perpendicular to $PQ$ and $P$ is any point, calculate $x$.(Answer:$1u^2$)

$\angle ANM \cong \angle NMC\\
\angle NMB \cong \angle MND$
$MN$ is common side
$$AB=CD$$
Is this enough to say that the two quadrilaterals are congruent?
 A: The right angles make the quad. PMCQ cyclic. Then, $\angle PMC= \angle PQD$. Therefore, quad. PMCQ and quad. PQDN are equi-angular.

In addition, 'the two quadrilaterals are equal in area' implies they are congruent.
Then, PM of quad. PMCQ = PQ of quad PQDN and PQ of quad. PMCQ = PN of quad. PQDN. Therefore, PM = PQ = PN. (*)
The same logic extends to the third.
The logic of (*) plus the fourth quadrilateral is equi-angular should be enough to show that those four quadrilaterals are equal in area.
A: Let $P'\in\overline{CD}$ be chosen so that $\overline{PP'}$ is parallel to $\overline{AD}$, and let $N'\in\overline{BC}$ be chosen so that $\overline{NN'}$ is parallel to $\overline{CD}$. Because $\overline{PQ}$ and $\overline{MN}$ are perpendicular, we see that $\angle QPP'\cong\angle MNN'$ and so $\triangle PP'Q\cong \triangle NN'M$. Thus, $P'Q = N'M$.
The areas of trapezoids $ADQP$ and $NDCM$ are equal, and so, since they have equal heights, $AP + DQ = ND + CM$. This means that $2AP + P'Q = 2ND + N'M$, and so $AP=ND$. Similarly, the areas of trapezoids $BPQC$ and $BANM$ are equal, and so $QC=MB$.  However, I cannot continue at this stage.
A: We can and do assume that the side of the square is one. Let us fix the position of $MN$, as shown in the picture below, and denote by $a,b$ the lengths of the segments $ND$ an $MC$.

We are searching first for a position of the segment $PQ$, with $PQ\perp MN$,
and

*

*with the equality of the areas $[PANR]$, and $[QCMR]$, denote by $U$ their common value,

*or equivalently (after adding $[RNDQ]$ to both) with the equality of the areas $[PADQ]$, and $[MNDC]$.

Since $[MNDC]$ is fixed, and the slide of a line $PQ$ "downwards",
parallel to itself on a direction perpendicular to $MN$, is a continuous movement cutting more and more from $[PADQ]$, the position of $PQ$ is unique. For short, the area $[PADQ]$ is a decreasing continuous function, seen as a function of $R\in[MN]$, which is moving from $M$ to $N$, and determining $PQ$. We know this position, it is realized by a $90^\circ$ rotation of $MN$ in the direction that brings the square $ABCD$ into $BCDA$, and $M$ in $Q$, $N$ in $P$, $MC=b$ in $QD=b$, $DN=a$ in $AP=a$.
So far we have arranged only that $U=[ANRP]=[CMRQ]$. Now we use the information that these areas are equal to $[DQRN]$. Again, fix the trapezium $APQD$ in sight, and let $R$ slide on $PQ$ from $P$ to $Q$, then draq the  perpendicular $RN$. It is clear that the area $[APRN]$ is a continuous, increasing function of $R$, so there exists a unique point $R$ realizing $[APRN]$ as the half of $[APQD]$. For this one point record the special value $AN:ND$. Now do the same with the trapezium $DNMC$, congruent with $APQD$. Consider again the point $R$ sliding on $NM$, it has by the same argument an existing unique position of $RQ$ to cut $[DNMC]$ in two equal areas, and the corresponding singular point $Q$ realizes the same proportion:
$$
\frac{AN}{ND}=
\frac{DQ}{QC}\ ,\qquad\text{ i.e. }\qquad
\frac{1-a}a=\frac{b}{1-b}\ .
$$
This gives $a=1-b$, so both $MN$ and $PQ$ are passing through the center of symmetry of the square. So the fourth piece $BPRM$ is congruent to the other three pieces, thus having the same area.
$\square$
