If in a triangle, the area is numerically equal to the perimeter, then the radius of the inscribed circle of the triangle is

### Concept & Inradius-Area Formula The fundamental relationship connecting the area of a triangle ($A$), its perimeter ($P$), and the radius of its inscribed circle (inradius $r$) is: $$A = r \times s$$ where $s$ is the semi-perimeter ($s = P / 2$). ### Step-by-Step Solution 1. **Express the Formula in Terms of Perimeter:** Substitute $s = \frac{P}{2}$ into the area formula: $A = r \times \frac{P}{2}$ 2. **Apply the Given Condition:** We are given that the area is numerically equal to the perimeter, so $A = P$. Substitute $A$ for $P$ in our equation: $A = r \times \frac{A}{2}$ 3. **Solve for $r$:** Divide both sides by $A$ (since area cannot be zero): $1 = \frac{r}{2}$ $r = 2$ ### Exam Strategy & Shortcut This is a theoretical property that can be memorized. Whenever a 2D shape's area is calculated as "radius times semi-perimeter" (which is true for all triangles and tangential polygons), if Area = Perimeter, the inradius is always exactly 2. ### Common Pitfall A common mistake is attempting to assume a specific type of triangle (like equilateral) and working through complex side-length formulas, which wastes time and often introduces calculation errors, when the relationship holds universally for all triangles. ### Final Answer Therefore, the correct answer is **2**.