The base of a straight prism is an isosceles trapezoid circumscribed around a circle of radius $r$, with an acute angle $\alpha$. The plane passing through one leg of the base and the opposite vertex of the acute angle of the other leg of the top face forms an angle $\alpha$ with the base plane. What is the height of the prism?
Let the endpoints of the oblique side of the base plate be denoted by $B$ and $C$ (with $B$ being the acute vertex), the acute vertex of the oblique side of the cover plate by $E$, the vertex of the base plate below $E$ by $A$, and the orthogonal projection of $A$ onto $BC$ by $T$. Since both $AT$ and $AE$ are perpendicular to the line $BC$, $BC$ is perpendicular to the plane $AET$. Therefore, the angle between the lines $AT$ and $ET$ is equal to the angle between the planes $ABC$ and $EBC$, and in the triangle $AET$,

$$
AE = AT \cdot \operatorname{tg} \alpha
$$

![](https://cdn.mathpix.com/cropped/2024_05_02_8f3023aa7602133a4829g-1.jpg?height=642&width=545&top_left_y=454&top_left_x=779)

The segment $AT$ represents the distance from the vertex $A$ of the base plate to the line $BC$. Let the center of the inscribed circle of the base plate be denoted by $O$, the projection of $O$ onto $BC$ by $U$, and the midpoint of $AB$ by $F$. The distance from $F$ to $BC$ is half of $AT$. This distance can be obtained by projecting $F$ onto the line $OU$ perpendicular to $BC$: if the projection is $V$, then $UV$ is the desired distance, and $AT = 2UV$. In the triangle $FOV$, $OF \perp FB$ and $OV \perp BU$, so $\angle FOV = \alpha$ and $OV = r \cos \alpha$. Therefore,

$$
AE = 2UV \cdot \operatorname{tg} \alpha = 2(OV + OU) \cdot \operatorname{tg} \alpha = 2(r \cos \alpha + r) \cdot \operatorname{tg} \alpha = 2r(\sin \alpha + \operatorname{tg} \alpha)
$$

![](https://cdn.mathpix.com/cropped/2024_05_02_8f3023aa7602133a4829g-1.jpg?height=524&width=469&top_left_y=1451&top_left_x=817)

Remark. In our solution, we assumed the isosceles trapezoid to be axis-symmetric, with the acute angles on the same parallel side. If we interpret the naming such that the quadrilateral is both a trapezoid and has equal legs, we can say that the quadrilateral could also be a rhombus. This is a somewhat forced interpretation, but if we interpret the problem statement this way, the result is $2r \operatorname{tg} \alpha$.