
	 Dodecagon $QWARTZSPHINX$ has all side lengths equal to $2$, is not self-intersecting (in particular, the twelve vertices are all distinct), and moreover each interior angle is either $90^{\circ}$ or $270^{\circ}$. What are all possible values of the area of $\triangle SIX$?
1. **Understanding the Problem:**
   - We have a dodecagon \(QWARTZSPHINX\) with all side lengths equal to 2.
   - The dodecagon is not self-intersecting, and each interior angle is either \(90^\circ\) or \(270^\circ\).
   - We need to find all possible values of the area of \(\triangle SIX\).

2. **Analyzing the Shape:**
   - Given the constraints, the dodecagon must form a "+" shape. This is because the only way to have all sides equal and the angles as specified is to have a shape that extends in perpendicular directions, forming a cross or "+" shape.

3. **Constructing the Dodecagon:**
   - The "+" shape can be visualized as a central square with four arms extending outwards. Each arm consists of two segments of length 2, making the total length of each arm 4.
   - The central square will have side lengths of 2, and the arms will extend outwards from the midpoints of the sides of the square.

4. **Identifying \(\triangle SIX\):**
   - To find the area of \(\triangle SIX\), we need to identify the vertices \(S\), \(I\), and \(X\) within the dodecagon.
   - Given the symmetry and structure of the "+" shape, there are three non-congruent possibilities for the placement of \(\triangle SIX\).

5. **Calculating the Area:**
   - **Case 1:**
     - If \(\triangle SIX\) is formed by three consecutive vertices on one of the arms, the area can be calculated using the formula for the area of a right triangle.
     - The base and height will each be 2 (since each side of the dodecagon is 2).
     - Area = \(\frac{1}{2} \times 2 \times 2 = 2\).

   - **Case 2:**
     - If \(\triangle SIX\) is formed by two vertices on one arm and one vertex on the central square, the area will be different.
     - The base will be 4 (the length of the arm), and the height will be 2 (the side of the central square).
     - Area = \(\frac{1}{2} \times 4 \times 2 = 4\).

   - **Case 3:**
     - If \(\triangle SIX\) is formed by three vertices on the central square, the area will be calculated similarly to Case 1.
     - The base and height will each be 2.
     - Area = \(\frac{1}{2} \times 2 \times 2 = 2\).

6. **Conclusion:**
   - The possible values of the area of \(\triangle SIX\) are 2 and 4.

The final answer is \(\boxed{2 \text{ and } 4}\)