A semicircle has diameter $XY$. A square $PQRS$ with side length 12 is inscribed in the semicircle with $P$ and $S$ on the diameter. Square $STUV$ has $T$ on $RS$, $U$ on the semicircle, and $V$ on $XY$. What is the area of $STUV$?
1. **Determine the radius of the semicircle:**
   - The side length of the square \(PQRS\) is 12, and it is inscribed in the semicircle with \(P\) and \(S\) on the diameter \(XY\).
   - Since \(PQRS\) is a square, the diagonal \(PR\) (or \(QS\)) is the hypotenuse of a right triangle with legs of length 12.
   - Using the Pythagorean theorem, the length of the diagonal \(PR\) is:
     \[
     PR = \sqrt{12^2 + 12^2} = \sqrt{144 + 144} = \sqrt{288} = 12\sqrt{2}
     \]
   - The diagonal \(PR\) is also the diameter of the semicircle, so the radius \(r\) of the semicircle is:
     \[
     r = \frac{12\sqrt{2}}{2} = 6\sqrt{2}
     \]

2. **Set up the system of equations for the square \(STUV\):**
   - Let \(x\) be the side length of the square \(STUV\).
   - The square \(STUV\) has \(T\) on \(RS\), \(U\) on the semicircle, and \(V\) on \(XY\).
   - The center of the semicircle is at the midpoint of \(XY\), which we can place at the origin \((0, 0)\) for simplicity.
   - The coordinates of \(S\) and \(T\) are \((6, 0)\) and \((6, x)\) respectively, since \(S\) is on the diameter and \(T\) is directly above \(S\).

3. **Find the coordinates of \(U\):**
   - Since \(U\) is on the semicircle, its coordinates must satisfy the equation of the semicircle:
     \[
     x^2 + y^2 = (6\sqrt{2})^2 = 72
     \]
   - The coordinates of \(U\) are \((6, y)\), where \(y\) is the height of the square above \(S\).
   - Since \(U\) is also a vertex of the square \(STUV\), the height \(y\) is equal to the side length \(x\) of the square.

4. **Solve for \(x\):**
   - Substitute \(y = x\) into the equation of the semicircle:
     \[
     6^2 + x^2 = 72
     \]
     \[
     36 + x^2 = 72
     \]
     \[
     x^2 = 36
     \]
     \[
     x = 6
     \]

5. **Calculate the area of the square \(STUV\):**
   - The side length of the square \(STUV\) is \(x = 6\).
   - The area of the square is:
     \[
     \text{Area} = x^2 = 6^2 = 36
     \]

The final answer is \(\boxed{36}\).