Given the square $ABCD$ with side $a$ and the square $BEFG$ with side $b$, we drew them next to each other as shown in the figure.

![](https://cdn.mathpix.com/cropped/2024_05_02_b89111ccd1c5831cbc8cg-1.jpg?height=293&width=453&top_left_y=225&top_left_x=825)

Express the area of the quadrilateral determined by the midpoints of segments $AB$, $BE$, $FC$, and $DG$ in terms of $a$ and $b$.
I. solution. Let $H, I, J$ and $K$ be the midpoints of segments $AB, BE, FC$ and $GD$. According to the problem statement:

$$
AB=BC=CD=AD=a \quad \text { and } \quad BE=EF=FG=BG=b
$$

![](https://cdn.mathpix.com/cropped/2024_05_02_225750f33a4b3cf78e56g-1.jpg?height=336&width=515&top_left_y=282&top_left_x=794)

We use the fact that the midline of a trapezoid (the segment connecting the midpoints of the non-parallel sides) is parallel to the bases and its length is equal to half the sum of the bases.

We know that $BG \| BC$, because this is the common side of the squares, $AD\|BC\|BG\|EF$, because these are the opposite sides of the respective squares.

Therefore, the quadrilaterals $ABGD$ and $EFBC$ are trapezoids where the bases are $AD=BC=a$ and $BG=EF=b$. Thus, the lengths of the midlines are equal:

$$
HK=IJ=\frac{a+b}{2}
$$

Since the midlines of the trapezoids are parallel to the bases, $\angle KHI = \angle JIH = 90^\circ$. Similarly, $AE\|DC\|$ and $FG$, and $DC=a$, $FG=b$, so $KJ$ is the midline of a trapezoid whose bases are $a$ and $b$ in length. Therefore,

$$
KJ=\frac{a+b}{2}, \quad KJ \| DC
$$

Thus, $\angle JKH = \angle KJI = 90^\circ$.

Since in the quadrilateral $HIJK$, $HI=IJ=JK=KH$, and the angles of the quadrilateral are $90^\circ$, this quadrilateral is a square, whose area is: $t=\left(\frac{a+b}{2}\right)^{2}$.

II. solution. Place the diagram in a coordinate system where the line $AE$ is the $x$-axis and the line $BC$ is the $y$-axis, so $B$ is the origin.

![](https://cdn.mathpix.com/cropped/2024_05_02_225750f33a4b3cf78e56g-1.jpg?height=357&width=553&top_left_y=1464&top_left_x=775)

The square $ABCD$ has side length $a$, so the coordinates of the vertices are: $A(-a; 0), B(0; 0), C(0; a), D(-a; a)$, where $a>0$.

The square $BEGF$ has side length $b$, so the coordinates of the vertices are: $B(0; 0), E(b; 0), F(b; b), G(0; b)$, where $b>0$.

Calculate the coordinates of the midpoints.

The midpoint of $AB$ is: $F_{1}\left(-\frac{a}{2}; 0\right)$, the midpoint of $BE$ is: $F_{2}\left(\frac{b}{2}; 0\right)$, the midpoint of $FC$ is: $F_{3}\left(\frac{b}{2}; \frac{a+b}{2}\right)$, the midpoint of $DG$ is: $F_{4}\left(-\frac{a}{2}; \frac{a+b}{2}\right)$. Then:

$$
F_{1}F_{2}=\sqrt{\left(-\frac{a}{2}-\frac{b}{2}\right)^{2}+(0-0)^{2}}=\frac{a+b}{2}
$$

Similarly, it can be calculated that:

$$
F_{2}F_{3}=F_{3}F_{4}=F_{4}F_{1}=\frac{a+b}{2}
$$

Thus, the quadrilateral $F_{1}F_{2}F_{3}F_{4}$ is a rhombus. Two side vectors of the rhombus are:

$$
\overrightarrow{F_{1}F_{2}}\left(\frac{a+b}{2}; 0\right), \quad \overrightarrow{F_{1}F_{4}}\left(0; \frac{a+b}{2}\right)
$$

Then $\overrightarrow{F_{1}F_{2}} \cdot \overrightarrow{F_{1}F_{4}}=0$, which means these vectors are perpendicular to each other. Therefore, the quadrilateral $F_{1}F_{2}F_{3}F_{4}$ is a square, whose side length is $\frac{a+b}{2}$. Based on this, the area in question is: $t=\left(\frac{a+b}{2}\right)^{2}$.