Find whether among all quadrilaterals, whose interiors lie inside a semi-circle of radius $r$, there exist one (or more) with maximum area. If so, determine their shape and area.
1. **Existence of Maximum Area Quadrilateral:**
   - The existence of a maximum area quadrilateral is guaranteed by the compactness argument. Since the semi-circle is a closed and bounded set, any continuous function (such as the area function) defined on it will attain its maximum value.

2. **Convexity and Vertices on Boundary:**
   - A quadrilateral achieving maximum area must be convex and have all its vertices on the boundary of the semi-circle. This is because a non-convex quadrilateral or one with vertices inside the semi-circle would have a smaller area compared to a convex quadrilateral with vertices on the boundary.

3. **Vertices on the Diameter:**
   - Assume two vertices lie on the diameter of the semi-circle. If not, consider the side with the least distance from the center of the circle of radius \( r \) and extend it to a full chord. This ensures the quadrilateral is inscribed in the semi-circle with diameter \( \leq 2r \).

4. **Symmetrization:**
   - Symmetrize the quadrilateral with respect to the diameter. The result is a polygon with 6 vertices, of double the area, contained in the circle of radius \( r \).

5. **Steiner's Result:**
   - By Steiner's result, the largest area of an \( n \)-gon contained in a circle is realized by the regular \( n \)-gon. Therefore, in our case, the maximum area is given by a half regular hexagon.

6. **Area Calculation:**
   - A regular hexagon inscribed in a circle of radius \( r \) has an area given by:
     \[
     A_{\text{hexagon}} = \frac{3\sqrt{3}}{2} r^2
     \]
   - Since we are considering a half regular hexagon, the area of the quadrilateral is:
     \[
     A_{\text{quadrilateral}} = \frac{1}{2} \times \frac{3\sqrt{3}}{2} r^2 = \frac{3\sqrt{3}}{4} r^2
     \]

The final answer is \( \boxed{ \frac{3\sqrt{3}}{4} r^2 } \)