1. **Problem statement:** We have a circle with center $O$ and radius $10$ cm. Two chords $AB$ and $PQ$ are parallel, with lengths $AB=12$ cm and $PQ=16$ cm. The line $OX$ is perpendicular to $PQ$. We need to find the distance between the chords $AB$ and $PQ$. 2. **Formula and rules:** The distance from the center of the circle to a chord can be found using the formula for the perpendicular distance $d$ from the center to the chord: $$d=\sqrt{r^2 - \left(\frac{c}{2}\right)^2}$$ where $r$ is the radius and $c$ is the chord length. 3. **Calculate distance from center to each chord:** - For chord $AB$: $$d_{AB}=\sqrt{10^2 - \left(\frac{12}{2}\right)^2}=\sqrt{100 - 36}=\sqrt{64}=8$$ - For chord $PQ$: $$d_{PQ}=\sqrt{10^2 - \left(\frac{16}{2}\right)^2}=\sqrt{100 - 64}=\sqrt{36}=6$$ 4. **Find the distance between chords:** Since $AB$ and $PQ$ are parallel and on the same side of the center (because $OX \perp PQ$ and $OX$ is radius), the distance between them is: $$\text{Distance} = d_{AB} - d_{PQ} = 8 - 6 = 2$$ **Final answer:** $$2$$