In an acute angle `Delta ABC,` let AD, BE and CF be the perpendicular from A, B and C upon the opposite sides of the triangle. (All symbols used have usual meaning in a tiangle.)
The circum-radius of the `Delta DEF` can be equal to

Options

A. `(abc)/(8Delta)`
B. `(a)/(4 sin A)`
C. `R/2`
D. `r/8 cosec ""A/2 cosec ""B/2 cosec ""C/2`

Answer

To solve the problem regarding the circum-radius of triangle DEF formed by the feet of the altitudes from vertices A, B, and C of triangle ABC, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Configuration**:
- Let triangle ABC be an acute triangle with vertices A, B, and C.
- Let D, E, and F be the feet of the perpendiculars dropped from A, B, and C to the opposite sides BC, AC, and AB, respectively.

2. **Identifying the Properties of Triangle DEF**:
- Triangle DEF is known as the orthic triangle of triangle ABC.
- The angles of triangle DEF can be expressed in terms of the angles of triangle ABC. Specifically, the angles of triangle DEF are related to the angles of triangle ABC as follows:
- ∠EDF = 90° - A
- ∠DEF = 90° - B
- ∠FDE = 90° - C

3. **Using the Sine Rule**:
- The circum-radius \( R' \) of triangle DEF can be expressed using the sine rule:
\[ \frac{EF}{\sin \angle EDF} = 2R' \]
- Since \( \angle EDF = 90° - A \), we have:
\[ \sin \angle EDF = \cos A \]
- Therefore, we can write:
\[ \frac{EF}{\cos A} = 2R' \]

4. **Finding the Length of EF**:
- The length of EF can be derived from the sides of triangle ABC. Using the cosine rule in triangle AFE:
\[ EF^2 = AF^2 + AE^2 - 2 \cdot AF \cdot AE \cdot \cos A \]
- From the properties of altitudes:
\[ AF = b \cos A \quad \text{and} \quad AE = c \cos B \]
- Substituting these into the equation gives:
\[ EF^2 = (b \cos A)^2 + (c \cos B)^2 - 2(b \cos A)(c \cos B) \cos A \]

5. **Simplifying the Expression**:
- After simplification, we find:
\[ EF = a \cos A \]
- Thus, substituting back into the sine rule equation:
\[ \frac{a \cos A}{\cos A} = 2R' \]
- This simplifies to:
\[ a = 2R' \]

6. **Relating R' to R**:
- From the triangle ABC, we know that:
\[ \frac{a}{\sin A} = 2R \]
- Therefore, we can equate:
\[ 2R' = 2R \cdot \frac{a}{a} \implies R' = \frac{R}{2} \]

### Conclusion:
The circum-radius \( R' \) of triangle DEF is:
\[ R' = \frac{R}{2} \]