The product of the perimeter of a triangle, the radius of its in-circle, and a number gives the area of the triangle. The number is

Correct option is C

Solution:

Consider a triangle ABC with incircle with radius r.

∴ OF = OH = OG = r.

Consider Δ ABO.

Here OF is perpendicular to AB (AB is tangent to the circle)

​$$\begin{aligned}&\text{Area of } \triangle ABO = \frac{1}{2} \times OF \times AB \\&\Rightarrow \text{Area of } \triangle ABO = \frac{1}{2} \times r \times AB \\[8pt]&\text{Similarly} \\[4pt]&\text{Area of } \triangle ACO = \frac{1}{2} \times OH \times AC \\&\Rightarrow \text{Area of } \triangle ACO = \frac{1}{2} \times r \times AC \\[8pt]&\text{Similarly} \\[4pt]&\text{Area of } \triangle BCO = \frac{1}{2} \times OG \times BC \\&\Rightarrow \text{Area of } \triangle BCO = \frac{1}{2} \times r \times BC \\[8pt]&\text{Area of } \triangle ABC = \text{Area of } \triangle ABO + \text{Area of } \triangle ACO + \text{Area of } \triangle BCO \\&\Rightarrow \text{Area of } \triangle ABC = \frac{1}{2} \times r \times AB + \frac{1}{2} \times r \times AC + \frac{1}{2} \times r \times BC \\&\Rightarrow \text{Area of } \triangle ABC = \frac{1}{2} \times r \times (AB + BC + AC) \\&\Rightarrow \text{Area of } \triangle ABC = \frac{1}{2} \times r \times \text{Semi-perimeter}\end{aligned}$$​​

​Hence the number is 1/2.

Final Answer: (C)