The area of the largest triangle that can be inscribed in a semi-circle whose radius is $r$ is:

$\textbf{(A)}\ r^2 \qquad
\textbf{(B)}\ r^3 \qquad
\textbf{(C)}\ 2r^2 \qquad
\textbf{(D)}\ 2r^3 \qquad
\textbf{(E)}\ \dfrac{1}{2}r^2$
1. Consider a semi-circle with radius \( r \). The diameter of the semi-circle is \( 2r \).
2. The largest triangle that can be inscribed in a semi-circle is a right-angled triangle with the hypotenuse as the diameter of the semi-circle. This is because the right angle will be subtended by the diameter, making it the largest possible triangle.
3. In a right-angled triangle inscribed in a semi-circle, the hypotenuse is the diameter of the semi-circle, which is \( 2r \).
4. For the largest area, the triangle should be an isosceles right-angled triangle (i.e., a \( 45^\circ - 45^\circ - 90^\circ \) triangle). In such a triangle, the legs are equal.
5. Let the length of each leg be \( a \). Using the Pythagorean theorem:
   \[
   a^2 + a^2 = (2r)^2
   \]
   \[
   2a^2 = 4r^2
   \]
   \[
   a^2 = 2r^2
   \]
   \[
   a = r\sqrt{2}
   \]
6. The area \( A \) of a right-angled triangle is given by:
   \[
   A = \frac{1}{2} \times \text{leg}_1 \times \text{leg}_2
   \]
   Substituting the values of the legs:
   \[
   A = \frac{1}{2} \times r\sqrt{2} \times r\sqrt{2}
   \]
   \[
   A = \frac{1}{2} \times 2r^2
   \]
   \[
   A = r^2
   \]

The final answer is \(\boxed{r^2}\).