28. Geometric problem. Given an ellipse, the length of the major axis is $2a$, and the length of the minor axis is $2b$. Draw a closed curve of the same length as the ellipse, enclosing an area that is greater than the area of the ellipse by $(a-b)^{2}$.

## CHAPTER III

## SPACE, POLYHEDRA, PLANES
28. The desired curve is obtained by dividing the ellipse into four parts and combining them as shown in Fig. 37.

Here is another solution that does not require "flipping" the parts, i.e., moving out of the plane of the ellipse. By connecting the vertices of the ellipse with chords alternately (Fig. 38), we obtain a rhombus surrounded by four elliptical segments. We replace this rhombus, whose area is $2 a b$, with a square whose area is
![](https://cdn.mathpix.com/cropped/2024_05_21_5d77f4829f88503a49eag-081.jpg?height=456&width=1108&top_left_y=620&top_left_x=470)

Fig. 37.

$c^{2}=a^{2}+b^{2}$, so that the four elliptical segments still adjoin the sides of the square. The area bounded by the curve increases by the difference in the areas of the square and the rhombus, i.e., by

$$
c^{2}-2 a b=a^{2}+b^{2}-2 a b=(a-b)^{2}
$$