In a `Delta ABC`, the `AM` and `GM` of ex-radii are 8 asnd 4 respectively, then inradius is :

Options

A. `3`
B. `2`
C. `1`
D. `(2)/(3)`

Explanation

To solve the problem, we need to find the inradius \( r \) of triangle \( ABC \) given that the arithmetic mean (AM) of the ex-radii is 8 and the geometric mean (GM) of the ex-radii is 4.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- AM of ex-radii \( = 8 \)
- GM of ex-radii \( = 4 \)

2. **Use the Relationship Between AM, GM, and Harmonic Mean (HM):**
- The relationship is given by the formula:
\[ G^2 = A \cdot H \]
- Here, \( G \) is the GM, \( A \) is the AM, and \( H \) is the HM.
- Substitute the known values:
\[ 4^2 = 8 \cdot H \]
- This simplifies to:
\[ 16 = 8H \]
- Solving for \( H \):
\[ H = \frac{16}{8} = 2 \]

3. **Relate HM to the Ex-Radii:**
- The harmonic mean \( H \) of the ex-radii \( r_1, r_2, r_3 \) can be expressed as:
\[ H = \frac{3}{\frac{1}{r_1} + \frac{1}{r_2} + \frac{1}{r_3}} \]
- Setting this equal to the value we found for \( H \):
\[ \frac{3}{\frac{1}{r_1} + \frac{1}{r_2} + \frac{1}{r_3}} = 2 \]

4. **Rearranging the Equation:**
- Cross-multiplying gives:
\[ 3 = 2 \left( \frac{1}{r_1} + \frac{1}{r_2} + \frac{1}{r_3} \right) \]
- Simplifying, we find:
\[ \frac{1}{r_1} + \frac{1}{r_2} + \frac{1}{r_3} = \frac{3}{2} \]

5. **Expressing the Sum of the Ex-Radii in Terms of the Semi-Perimeter:**
- The ex-radii can be expressed in terms of the area \( \Delta \) and semi-perimeter \( s \):
\[ r_1 = \frac{\Delta}{s-a}, \quad r_2 = \frac{\Delta}{s-b}, \quad r_3 = \frac{\Delta}{s-c} \]
- Thus,
\[ \frac{1}{r_1} + \frac{1}{r_2} + \frac{1}{r_3} = \frac{s-a}{\Delta} + \frac{s-b}{\Delta} + \frac{s-c}{\Delta} \]
- This simplifies to:
\[ \frac{(s-a) + (s-b) + (s-c)}{\Delta} = \frac{3s - (a+b+c)}{\Delta} = \frac{3s - 2s}{\Delta} = \frac{s}{\Delta} \]

6. **Setting the Equation Equal to the Previously Found Value:**
- We have:
\[ \frac{s}{\Delta} = \frac{3}{2} \]
- Rearranging gives:
\[ \Delta = \frac{2s}{3} \]

7. **Finding the Inradius \( r \):**
- The inradius \( r \) is given by:
\[ r = \frac{\Delta}{s} \]
- Substituting the value of \( \Delta \):
\[ r = \frac{\frac{2s}{3}}{s} = \frac{2}{3} \]

### Final Answer:
The inradius \( r \) is \( \frac{2}{3} \).