A rhombus has diagonal lengths of 6 and 8 units. A regular triangle is inscribed in the rhombus, with one vertex at one endpoint of the shorter diagonal, and one side parallel to the longer diagonal of the rhombus. What is the height of this triangle?
Let the vertices of the rhombus be $A, B, C$, and $D$, with $A C$ being the shorter diagonal and $B D$ the longer diagonal. Let $A$ be one vertex of the inscribed equilateral triangle. The other two vertices of the equilateral triangle, $E$ and $F$, lie on the sides $D C$ and $C B$ of the rhombus, respectively.

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Since $E F \| D B$ and $A C \perp D B$, the diagonals of the rhombus are perpendicular to each other and bisect each other. Therefore, $E F \perp A C$. Thus, $A C$ is the altitude from $A$ in the equilateral triangle. Let the intersection of the diagonals be $P$, and the midpoint of segment $E F$ be $Q$, which is also the foot of the altitude from $A$. The desired altitude is $A Q$.

From the similarity of triangles $E Q C$ and $D P C$,

$$
\frac{E Q}{Q C}=\frac{D P}{P C}=\frac{4}{3}
$$

The altitude of an equilateral triangle is $\sqrt{3} / 2$ times its side length, so $A Q=(\sqrt{3} / 2) E F=\sqrt{3} E Q$. Thus, from (1),

$$
\frac{4}{3}=\frac{E Q}{Q C}=\frac{A Q / \sqrt{3}}{A C-A Q}=\frac{A Q}{\sqrt{3}(6-A Q)}
$$

From this, the desired altitude is $A Q=\frac{24}{\sqrt{3}+4}=c \frac{96-24 \sqrt{3}}{13} \approx 4.187$ units. This is the altitude of the inscribed equilateral triangle.

Remarks. 1. One polygon is inscribed in another if the vertices of the latter lie on the sides of the former.

2. It was unnecessary to specify that one side of the equilateral triangle should be parallel to the longer diagonal. Indeed, if we rotate the rhombus $60^{\circ}$ counterclockwise around vertex $A$, it is clear that only one side of the rotated rhombus intersects the original rhombus, specifically the side $B^{\prime} C^{\prime}$, which intersects side $B D$. Since an equilateral triangle can be inscribed in the rhombus such that one of its sides is parallel to the diagonal, it follows that this is the only way to inscribe it.