Figure Q1 shows an equilateral triangle OPQ, of side 12 cm, and the point S such that OS = PS = QS. The arc PXQ has center O and radius 12 cm. Find the perimeter of the shaded region, giving your answer in terms of π and √3.

The problem describes an equilateral triangle OPQ with side length 12 cm. The arc PXQ has center O and radius 12 cm. This means P and Q are points on the arc, and since OP = OQ = 12 cm (sides of the equilateral triangle), the angle $\angle POQ$ is $60^\circ$. Point S is defined such that OS = PS = QS, meaning S is the circumcenter of $\triangle OPQ$. Given the requirement for the answer to be in terms of $\pi$ and $\sqrt{3}$, the most consistent interpretation of the "shaded region" is the area bounded by the arc PXQ and the line segments PS and QS.

The arc PXQ has center O and radius $r = 12$ cm.
Since $\triangle OPQ$ is equilateral, the angle $\angle POQ$ subtended by the arc at the center O is $60^\circ$.
Convert this angle to radians:

The length of an arc is given by the formula $L = r\theta$.

S is the circumcenter of the equilateral triangle OPQ. The distance from any vertex of an equilateral triangle to its circumcenter is the circumradius, $R$.
For an equilateral triangle with side length $a$, the circumradius is given by $R = \frac{a}{\sqrt{3}}$.
Here, the side length $a = 12$ cm.

To rationalize the denominator, multiply the numerator and denominator by $\sqrt{3}$:

Therefore, the lengths of segments PS and QS are both $4\sqrt{3}$ cm.

The perimeter of the shaded region is the sum of the arc length and the lengths of the two segments PS and QS.

The perimeter of the shaded region is $*(4\pi + 8\sqrt{3}) \text{ cm}*$.