6. Let the perimeter of $\triangle A B C$ be 12, and the radius of the inscribed circle be 1. Then ( ).
(A) $\triangle A B C$ must be a right triangle
(B) $\triangle A B C$ must be an acute triangle
(C) $\triangle A B C$ must be a right triangle or an acute triangle
(D) None of the above conclusions is correct
6. D.

Since the perimeter of $\triangle ABC$ is 12, the inradius of $\triangle ABC$ is 1 if and only if the area of $\triangle ABC$ is $\frac{12}{2} \times 1 = 6$.
From equation (2), we get
$a^{2} + b^{2} - 2ab \cos C$
$= a^{2} + b^{2} + 144 + 2ab - 24a - 24b$
$\Rightarrow a + b = 6 + \frac{1}{12} ab(1 + \cos C)$.
From equation (1), we get $ab = \frac{12}{\sin C}$, substituting this into the above equation, we get
$a + b = 6 + \frac{1 + \cos C}{\sin C}$.
Thus, $a$ and $b$ are the roots of the equation
$$
x^{2} - \left(6 + \frac{1 + \cos C}{\sin C}\right) x + \frac{12}{\sin C} = 0
$$

In particular, when $\angle C = 90^{\circ}$, we solve to get $\{a, b\} = \{3, 4\}$. At this time, $c = 5$, and the discriminant of equation (3) is
$$
\Delta = 7^{2} - 4 \times 12 = 1.
$$

Furthermore, when $\angle C$ increases by a very small angle from $90^{\circ}$, the discriminant $\Delta$ of the equation remains greater than 0, at this time, we can still solve for $a$ and $b$ from the system of equations, and then get $c$, at this time, the side lengths are also very close to 3, 4, and 5.

Therefore, there exists an obtuse triangle with a perimeter of 12 and an inradius of 1.