Let $K$ and $M$ be the midpoints of the diagonals $AC$ and $BD$ of trapezoid $ABCD$, and $L$ and $N$ be the midpoints of the bases $BC$ and $AD$, respectively, with $AB \perp CD$, $AB = a$, and $CD = b$.

Segments $KL$ and $MN$ are the midlines of triangles $ABC$ and $ABD$ with the common base $AB$, so $KL \| AB$, $KL = \frac{1}{2} AB = \frac{1}{2} a$, $MN \| AB$, $MN = \frac{1}{2} AB = \frac{1}{2} a$. Similarly, $KN = LM = \frac{1}{2} CD = \frac{1}{2} b$. The opposite sides of quadrilateral $KLMN$ are equal, so it is a parallelogram. Since its sides are parallel to the legs $AB$ and $CD$ of trapezoid $ABCD$, $KLMN$ is a rectangle. Its area is the product of adjacent sides, where $KL = \frac{1}{2} a$ and $LM = \frac{1}{2} b$. Therefore,

$$
S_{KLMN} = KL \cdot LM = \frac{1}{2} AB \cdot \frac{1}{2} CD = \frac{1}{2} a \cdot \frac{1}{2} b = \frac{1}{4} ab
$$
[Varignon Parallelogram] [Area of a Quadrilateral] The lateral sides of the trapezoid lie on perpendicular lines. Find the area of the quadrilateral with vertices at the midpoints of the diagonals and the midpoints of the bases of the trapezoid, if its lateral sides are equal to $a$ and $b$.