Let's construct a right tangential trapezoid, given the radius of the inscribed circle and the perimeter of the trapezoid.
I. solution. We know that if a circle can be inscribed in a right-angled trapezoid, then the diameter of the circle is equal to the length of the right side, $AB$, which is also the distance between the parallel sides, $s$.

Let the points of tangency of the circle with the sides $AB, BC, CD, DA$ be denoted by $E, F, G$, and $H$, respectively. We know that the lengths of the tangent segments from an external point to a circle are equal, i.e., $AE = AH = r$, $BE = BF = r$ (where $r$ is the radius of the inscribed circle), and $CF = CG$ and $DG = DH$. Therefore, if we subtract $AE + AH + BE + BF = 4r$ from the given perimeter $s$ and divide the remainder (if any) by 2, we get the length of the side $CD$ of the trapezoid. This implies that the problem has a solution if $s - 4r > 0$, and more specifically, if $\frac{s - 4r}{2} \geq 2r = CD$, i.e., $s > 8r$. In the case of equality, $CD = 2r$, and the trapezoid is a square.
![](https://cdn.mathpix.com/cropped/2024_05_02_eeb653d9ba34619d9c7eg-1.jpg?height=1184&width=740&top_left_y=516&top_left_x=692)

The construction proceeds as follows. We draw two parallel lines $e$ and $f$ that are $2r$ units apart. We take a point $A$ on $e$. The perpendicular to $e$ at $A$ intersects $f$ at $B$. Then we draw a circle that is tangent to all three lines. Next, we construct the length $CD = \frac{s}{2} - 2r$ on a segment of length $s$. Now we need to construct a tangent to the circle of length $CD$. This can be done, for example, by drawing a line from $B$ at a distance $CD$ to intersect $e$, and then drawing a perpendicular from the center of the circle to this line. This intersects the circle at point $G$, which determines the position of $CD$.

If the problem has a solution, it is clearly unique.

II. solution. Notice that the triangle $COD$ is a right triangle. Since $\angle BCD + \angle CDA = 180^\circ$, and $O$ is the intersection of the angle bisectors of these two angles. Based on this, the construction proceeds as follows: according to the first solution, $CD = \frac{s - 4r}{2}$, we first construct this distance. Then, given $CD$ and $r$, we construct a right triangle with hypotenuse $CD$ and height $r$. This can easily be extended to a right-angled trapezoid with an inscribed circle of radius $r$.