In a triangle, the area is numerically equal to the perimeter. What is the radius of the inscribed circle?

$\textbf{(A) }2\qquad\textbf{(B) }3\qquad\textbf{(C) }4\qquad\textbf{(D) }5\qquad \textbf{(E) }6$
1. Let \( a, b, c \) be the side lengths of the triangle. Let \( A \) be the area of the triangle. Let \( s = \frac{a+b+c}{2} \) be the semi-perimeter of the triangle.
2. We know that the area \( A \) of the triangle can be expressed in terms of the inradius \( r \) and the semi-perimeter \( s \) as:
   \[
   A = rs
   \]
3. From the problem, we are given that the area \( A \) is numerically equal to the perimeter \( a + b + c \). Therefore, we have:
   \[
   A = a + b + c
   \]
4. Using the formula for the inradius \( r \), we can write:
   \[
   r = \frac{A}{s}
   \]
5. Substituting \( s = \frac{a+b+c}{2} \) into the equation, we get:
   \[
   r = \frac{A}{\frac{a+b+c}{2}} = \frac{2A}{a+b+c}
   \]
6. Since \( A = a + b + c \) from the problem statement, we substitute \( A \) with \( a + b + c \):
   \[
   r = \frac{2(a+b+c)}{a+b+c} = 2
   \]

The final answer is \(\boxed{2}\)