Solve the following geometry problems involving circles and chords.

Since $U$ is the center and $UQ \perp PR$, $Q$ is the midpoint of chord $PR$. (Reason: Line from center perpendicular to chord bisects the chord).

In $\triangle UQP$, we have a right-angled triangle at $Q$.

Assuming $UA$ refers to the radius of the circle (since $U$ is the center and $A$ is not defined as a specific point on the circumference), then $UA = UP$.
The length of $UA$ is $\boxed{\text{5 cm}}$.

$TR$ is the diameter, and $U$ is the center. $UP$ is the radius.

Since $TR$ is the diameter, the angle subtended by the diameter at any point on the circumference is $90^\circ$. (Reason: Angle in a semicircle is $90^\circ$). Therefore, $\angle TSR = 90^\circ$.

Divide the entire equation by 2:

Factor the quadratic equation:
We need two numbers that multiply to -48 and add to 2. These numbers are 8 and -6.

So, $x+8=0$ or $x-6=0$.

Since length cannot be negative, we discard $x = -8$.
Therefore, $x = 6 \text{ cm}$.

$TS = x \text{ cm}$.

The length of $TS$ is $\boxed{\text{6 cm}}$.