6. Let's introduce the following notations: $a$ - the side of the base, $b$ - the lateral edge, $N$ - the midpoint of side $AB$, $M$ - the midpoint of side $AD$, $K$ - the midpoint of edge $SC$ of the regular quadrilateral pyramid $SABCD$.

![](https://cdn.mathpix.com/cropped/2024_05_06_f36d8ea35c99389d52e3g-05.jpg?height=1057&width=680&top_left_y=1156&top_left_x=745)

We construct the section of the pyramid $SABCD$ by a plane passing through points $N$ and $M$ and parallel to the median $DK$ of the lateral face $SDC$: through points $N$ and $M$ we draw the line $NM$ until it intersects with lines $BC$ and $CD$ at points $N_1$ and $M_1$ respectively; through point $M_1$ we draw a line parallel to the median $DK$, the points of intersection of this line with edges $SD$ and $SC$ are denoted as $P$ and $R$ respectively; we connect points $R$ and $N_1$; the point of intersection of $RN_1$ and $SB$ is denoted as $Q$; we connect $Q$ and $N$, $P$ and $M$. The section $NMPRQ$ is constructed. Let's find its area.

By the condition, $AM = MD = AN = NB = \frac{a}{2}$. Therefore, triangle $AMN$ is a right isosceles triangle and $MN = \frac{a}{\sqrt{2}}$. Note that triangles $MDM_1$ and $NBN_1$ are right triangles. They are equal to triangle $AMN$ by the leg and angle. Therefore, $M_1D = N_1B = \frac{a}{2}$. Consider the plane of the lateral face $SDC$. Since $M_1R \parallel DK$, then $\frac{RK}{CK} = \frac{M_1D}{CD} = \frac{1}{2}$. Then $RK = \frac{CK}{2} = \frac{b}{4}$, $SR = SK - RK = \frac{b}{4}$. Moreover, $PR \parallel DK$, $\frac{SP}{PD} = \frac{SR}{RK} = 1$. Therefore, $SP = PD = \frac{b}{2}$. Since $P$ is the midpoint of edge $SD$ and $M$ is the midpoint of base $AD$, then $PM$ is the midline of triangle $ADS$ and $PM = \frac{AS}{2} = \frac{b}{2}$, $PM \parallel AS$. Similarly, we can obtain that $QN = \frac{AS}{2} = \frac{b}{2}$, $QN \parallel AS$. Therefore, quadrilateral $PMQN$ is a parallelogram. We will prove that it is a rectangle. Indeed, the projection of $AC$ of edge $AS$ is the diagonal of the square $ABCD$. Therefore, $AC$ is perpendicular to the diagonal $BD$, and since $BD \parallel MN$, then $AC$ is perpendicular to $MN$. Hence, $AS$ is perpendicular to $MN$. Since $PM \parallel AS$, it follows that $PM$ is perpendicular to $MN$. Thus, the section $NMPRQ$ consists of a rectangle $PMQN$ with sides $MN = \frac{a}{\sqrt{2}}$ and $PM = \frac{b}{2}$ and an isosceles triangle $PQR$ with base $PQ = \frac{b}{2}$ and height $RF$. Let's find the area of the rectangle: $MN \cdot PM = \frac{ab}{2\sqrt{2}}$. Since $\frac{CR}{RS} = \frac{CO}{OA} = 3$, then $\frac{OR}{AS} = \frac{CO}{OA} = \frac{3}{4}$. From this, we get $OR = \frac{3b}{4}$, and $FR = \frac{3b}{4} - \frac{b}{2} = \frac{b}{4}$. Then the area of triangle $PQR$ is $\frac{1}{2} \cdot \frac{a}{\sqrt{2}} \cdot \frac{b}{4} = \frac{ab}{8\sqrt{2}}$. The area of the section $NMPRQ$ is: $\frac{ab}{2\sqrt{2}} + \frac{ab}{8\sqrt{2}} = \frac{5ab}{8\sqrt{2}} = \frac{5ab\sqrt{2}}{16}$. Substituting $a = 1$, $b = 2$, we find $S = \frac{5\sqrt{2}}{8}$.
Answer: $S=\frac{5 \sqrt{2}}{8}$.

#