In the figure below, PQRS is a rhombus attached to two semicircles and length QR is twice UV. The area of the rhombus is 24cm². Study and use it to answer the questions that follow. Find the length; (i) PQ

From the diagram, U is the intersection of the diagonals QS and PR. In a rhombus, the diagonals bisect each other.
Given that $US = 3 \text{ cm}$.
Therefore, $QU = US = 3 \text{ cm}$.
The full length of diagonal $QS = QU + US = 3 \text{ cm} + 3 \text{ cm} = 6 \text{ cm}$.

The area of a rhombus is given by the formula $A = \frac{1}{2} d_1 d_2$, where $d_1$ and $d_2$ are the lengths of the diagonals.
Given Area $A = 24 \text{ cm}^2$ and $d_1 = QS = 6 \text{ cm}$.

Since U is the midpoint of PR, $PU = \frac{1}{2} PR = \frac{1}{2} \times 8 \text{ cm} = 4 \text{ cm}$.

The diagonals of a rhombus intersect at right angles. Thus, $\triangle PUQ$ is a right-angled triangle with the right angle at U.
Using the Pythagorean theorem ($a^2 + b^2 = c^2$):

The length of PQ is $\boxed{5 \text{ cm}}$.

Since PQRS is a rhombus, all its sides are equal in length.
Therefore, $QR = PQ$.
From Part (i), we found $PQ = 5 \text{ cm}$.
So, $QR = 5 \text{ cm}$.

The problem states that "length QR is twice UV".

Substitute the value of QR:

The length of UV is $\boxed{2.5 \text{ cm}}$.

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