Let's construct a rhombus, given its perimeter: $k$, and the sum of its diagonals: $u(=e+f)$. Determine the solvability conditions.
In Figure 1, we have drawn the supposed rhombus $ABCD$.

![](https://cdn.mathpix.com/cropped/2024_05_02_5e190bc685d71d0adb97g-1.jpg?height=453&width=656&top_left_y=234&top_left_x=710)

One side of the rhombus and the right-angled triangle $AEB$ formed by the diagonals have a known hypotenuse: $\frac{k}{4}$, and the sum of the two legs is known: $\frac{e+f}{2}=\frac{u}{2}$. The question is, how can the right-angled triangle $AEB$ be constructed from this.

Measure, for example, the shorter half-diagonal, $EB$, on the extension of the longer one. In the resulting triangle $AB'B$, two sides are known: $\frac{k}{4}$ and $\frac{u}{2}$, and $\angle B' = 45^\circ$.

The construction can be completed most simply in the following way (Figure 2): On a segment $AB' = \frac{u}{2}$, measure a $45^\circ$ angle at point $B'$, then intersect the other ray of the angle with a circle of radius $\frac{k}{4}$ centered at point $A$, and draw a perpendicular from the intersection point $B$ (and $B_1$) to $AB'$. The resulting triangles $ABE$ and $AB_1E_1$ meet the requirements of the problem. The vertices of the rhombus can then be obtained, for example, by reflecting the other two vertices of the triangle over point $E$.

![](https://cdn.mathpix.com/cropped/2024_05_02_5e190bc685d71d0adb97g-1.jpg?height=656&width=730&top_left_y=1323&top_left_x=684)

Figure 2

It only appears that we have obtained two solutions, but it can be shown that the two are congruent. For example, draw a perpendicular to $AB'$ at point $A$, which intersects the line $B'B$ at a point $B_1'$ such that $AB_1' = AB'$. From point $B_1$ draw a perpendicular to $AB_1'$, and the triangles $ABE$ and $AB_1E_1'$ are mirror images of each other with respect to the symmetry axis of triangle $AB'B_1'$. Triangles $AB_1E_1'$ and $AB_1E_1$ are congruent.

The condition for the solvability of the problem is that the circle centered at $A$ has a common point with the segment $B'B_1'$ other than the endpoints, i.e.,

$$
AB' > AB \geq \frac{\sqrt{2}}{2} AB', \quad \text{so} \quad \frac{u}{2} > \frac{k}{4} \geq \frac{\sqrt{2} u}{4}
$$

or

$$
\sqrt{2} u \leq k < 2 u
$$

In the case $k = \sqrt{2} u$, we get a special rhombus: a square. In the case $k = 2 u$, the rhombus collapses into a line segment.