For a triangle `ABC, R = (5)/(2) and r = 1`. Let D, E and F be the feet of the perpendiculars from incentre I to BC, CA and AB, respectively. Then the value of `((IA) (IB)(IC))/((ID)(IE)(IF))` is equal to _____

Explanation

To solve the problem, we need to find the value of \(\frac{IA \cdot IB \cdot IC}{ID \cdot IE \cdot IF}\) for triangle \(ABC\) with given circumradius \(R = \frac{5}{2}\) and inradius \(r = 1\).

### Step-by-Step Solution:

1. **Understanding the terms**:
- \(IA\), \(IB\), and \(IC\) are the distances from the incenter \(I\) to the vertices \(A\), \(B\), and \(C\) respectively.
- \(ID\), \(IE\), and \(IF\) are the distances from the incenter \(I\) to the sides \(BC\), \(CA\), and \(AB\) respectively.

2. **Using the formulas for distances**:
- The distances from the incenter to the vertices can be expressed as:
\[ IA = R \cos \frac{A}{2}, \quad IB = R \cos \frac{B}{2}, \quad IC = R \cos \frac{C}{2} \]
- The distances from the incenter to the sides can be expressed as:
\[ ID = r, \quad IE = r, \quad IF = r \]
- Therefore, we have:
\[ ID \cdot IE \cdot IF = r^3 \]

3. **Substituting the values**:
- Now substituting the values into the expression:
\[ \frac{IA \cdot IB \cdot IC}{ID \cdot IE \cdot IF} = \frac{(R \cos \frac{A}{2})(R \cos \frac{B}{2})(R \cos \frac{C}{2})}{r^3} \]
- This simplifies to:
\[ \frac{R^3 \cdot \cos \frac{A}{2} \cdot \cos \frac{B}{2} \cdot \cos \frac{C}{2}}{r^3} \]

4. **Substituting the known values of \(R\) and \(r\)**:
- We know \(R = \frac{5}{2}\) and \(r = 1\), so:
\[ \frac{R^3}{r^3} = \frac{\left(\frac{5}{2}\right)^3}{1^3} = \frac{\frac{125}{8}}{1} = \frac{125}{8} \]

5. **Finding \(\cos \frac{A}{2} \cdot \cos \frac{B}{2} \cdot \cos \frac{C}{2}\)**:
- Using the identity:
\[ \cos \frac{A}{2} \cdot \cos \frac{B}{2} \cdot \cos \frac{C}{2} = \frac{1}{4} \cdot \frac{r}{R} \]
- Substituting \(r = 1\) and \(R = \frac{5}{2}\):
\[ \cos \frac{A}{2} \cdot \cos \frac{B}{2} \cdot \cos \frac{C}{2} = \frac{1}{4} \cdot \frac{1}{\frac{5}{2}} = \frac{1}{4} \cdot \frac{2}{5} = \frac{1}{10} \]

6. **Final calculation**:
- Now substituting back into the expression:
\[ \frac{IA \cdot IB \cdot IC}{ID \cdot IE \cdot IF} = \frac{125}{8} \cdot \frac{1}{10} = \frac{125}{80} = \frac{25}{16} \]

### Final Answer:
The value of \(\frac{IA \cdot IB \cdot IC}{ID \cdot IE \cdot IF}\) is \(\frac{25}{16}\).