In the figure below, PQRT and PRSU are rectangles. PQ = 8 cm, QR = 6 cm and PR = 10 cm. Find the length of RS.

1. Find the area of triangle PRT.

In rectangle $PQRT$, the diagonal $PR$ divides it into two congruent right-angled triangles.
Area of $\triangle PRT = \frac{1}{2} \times PT \times RT$
Since $PT = QR = 6$ cm and $RT = PQ = 8$ cm:
Area of $\triangle PRT = \frac{1}{2} \times 6 \times 8 = 24 \text{ cm}^2$

2. Relate the area of triangle PRT to the rectangle PRSU.

The distance from $T$ to the line $PR$ is the height of $\triangle PRT$ when $PR$ is taken as the base.
Let this height be $h$.
$\frac{1}{2} \times PR \times h = 24$
$\frac{1}{2} \times 10 \times h = 24$
$5h = 24 \implies h = 4.8 \text{ cm}$

3. Determine the length of RS.

In the rectangle $PRSU$, the side $RS$ is perpendicular to $PR$. The vertex $T$ lies on the opposite side $US$ (or the line containing $US$), meaning the perpendicular distance from $T$ to $PR$ is exactly the width of the rectangle $PRSU$, which is $RS$.
Therefore, $RS = h = 4.8 \text{ cm}$.