**PARAGRAPH \"I\"**\nConsider an obtuse angled triangle $ABC$ in which the difference between the largest and the smallest angle is $\frac{\pi}{2}$ and whose sides are in arithmetic progression. Suppose that the vertices of this triangle lie on a circle of radius $1$.\n\nThen the inradius of the triangle $ABC$ is
Step-by-Step Solution
Key Concept: Solving a homogeneous differential equation and finding the specific solution given a condition.
**Step 1: Recall the area and side relations**
From the previous problem, the area of the triangle is $\Delta = \frac{3\sqrt{7}}{16}$.\nThe side $b = \frac{\sqrt{7}}{2}$.\nThe sides $a, b, c$ are in arithmetic progression, so $a + c = 2b$.
**Step 2: Calculate the semi-perimeter**
The semi-perimeter $s$ is given by $s = \frac{a + b + c}{2}$.\nSubstituting $a + c = 2b$, we get $s = \frac{3b}{2}$.\nUsing the value of $b$, $s = \frac{3}{2} \left( \frac{\sqrt{7}}{2} \right) = \frac{3\sqrt{7}}{4}$.
**Step 3: Calculate the inradius**
The inradius $r$ is related to the area and semi-perimeter by $r = \frac{\Delta}{s}$.\n$r = \frac{\frac{3\sqrt{7}}{16}}{\frac{3\sqrt{7}}{4}} = \frac{3\sqrt{7}}{16} \times \frac{4}{3\sqrt{7}} = \frac{4}{16} = \frac{1}{4} = 0.25$.\nThe value is exactly $0.25$.
Correct Answer: 0.25