1. **Problem statement:** In rectangle MNPQ, one vertex lies on the circle, and two vertices lie on the diameter of the circle. Given $MN=4$ and $MQ=6$, find $QL$. 2. **Understanding the problem:** We have a rectangle inscribed in a circle such that one vertex is on the circle and two vertices lie on the diameter. We know two side lengths and need to find the length $QL$. 3. **Key properties:** In a rectangle inscribed in a circle, the diagonal equals the diameter of the circle. Also, opposite sides are equal and adjacent sides are perpendicular. 4. **Given:** $MN=4$, $MQ=6$. Since $MN$ and $MQ$ share vertex $M$, they are adjacent sides of the rectangle. 5. **Find:** $QL$, which is the side opposite to $MN$ or $MQ$ depending on labeling. 6. **Step:** Since $MN$ and $MQ$ are adjacent sides, the diagonal $PQ$ can be found by Pythagoras theorem: $$PQ=\sqrt{MN^2 + MQ^2} = \sqrt{4^2 + 6^2} = \sqrt{16 + 36} = \sqrt{52} = 2\sqrt{13}$$ 7. **Step:** Because $PQ$ is the diagonal, and $QL$ is a side of the rectangle adjacent to $PQ$, and $QL$ is equal to $MN$ or $MQ$ depending on the rectangle's orientation. But since $QL$ is not given, and $PQ$ is the diagonal, $QL$ must be equal to $MN=4$ or $MQ=6$. 8. **Step:** However, the problem asks to find $QL$, which is the chord length opposite to $MQ$. Since $QL$ is the chord subtending the same arc as $MN$, by the property of chords in a circle, $QL = MN = 4$. **Final answer:** $$QL=4$$