The image you've provided shows a rhombus with vertices $P$, $Q$, $R$, and $S$. You are asked to find the measure of the indicated side $PS$, and you are given that the diagonal $RS$ is 12 units long. In a rhombus, all sides are equal, and the diagonals bisect each other at right angles.

Step-by-step solution:

1. In a rhombus, the diagonals bisect each other perpendicularly (at 90 degrees) and divide the rhombus into four congruent right triangles.
2. The diagonal $RS = 12$. Since diagonals bisect each other, half of $RS$ is $\frac{12}{2} = 6$. Therefore, $RS$ is split into two segments of 6 units each.
3. The diagonals in a rhombus are also perpendicular to each other, so triangle $PRS$ is a right triangle.
4. In triangle $PRS$, $PR = PQ = PS = QR$ (since all sides of a rhombus are equal), and you can use the Pythagorean theorem to find $PS$.
5. Let the length of each side (including $PS$) be $x$. Applying the Pythagorean theorem to the right triangle $PRS$:

$x^2 = \left(\frac{RS}{2}\right)^2 + \left(\frac{PQ}{2}\right)^2$

$x^2 = 6^2 + 6^2$

$x^2 = 36 + 36 = 72$

$x = \sqrt{72} = 6\sqrt{2}$

Thus, the length of side $PS$ is $6\sqrt{2}$ units.

Would you like further details or have any questions about the explanation?

Here are some related questions to expand the concept:

1. What is the formula for the area of a rhombus using its diagonals?
2. How do you find the length of the diagonals if only the side lengths of a rhombus are given?
3. What is the relationship between the angles formed by the diagonals in a rhombus?
4. How do you calculate the perimeter of a rhombus given one diagonal?
5. Can a square be considered a special type of rhombus? Why?

Tip: In any rhombus, the diagonals always bisect each other at right angles, which helps in applying the Pythagorean theorem for finding side lengths.