In a parallelogram, one of the parallel sides is $16\text{cm}$ and the other side is $12\text{cm}$. If the perpendicular distance between the two parallel sides of dimension $16\text{cm}$ is $24\text{cm}$, then the perpendicular distance between its other two parallel sides is:

The area of a parallelogram can be computed in two ways using different bases and their corresponding perpendicular heights:
$$\text{Area}={\text{Base}}_1\times {\text{Height}}_1={\text{Base}}_2\times {\text{Height}}_2$$
Given:${\text{Base}}_1=16\text{cm}$ and its corresponding height ${\text{Height}}_1=24\text{cm}$${\text{Base}}_2=12\text{cm}$ and let its corresponding height be ${\text{Height}}_2=h$ Equating the areas:
$$16\times 24=12\times h$$
Solving for $h$:
$$h=\frac{16\times 24}{12}$$
$$h=16\times 2=32\text{cm}$$
Therefore, the perpendicular distance between the other two parallel sides is $32\text{cm}$.