If the perimeter of the triangle formed by feet of altitudes of the triangle ABC is equal to four times the circumradius of `Delta ABC`, then identify the type of `Delta ABC`

Explanation

To solve the problem, we need to analyze the given information about triangle ABC and the triangle formed by the feet of the altitudes of triangle ABC.

### Step-by-Step Solution:

1. **Understanding the Problem**: We are given that the perimeter of the triangle formed by the feet of the altitudes of triangle ABC (let's call it triangle DEF) is equal to 4 times the circumradius (R) of triangle ABC. We need to identify the type of triangle ABC.

2. **Setting Up the Equation**: The perimeter of triangle DEF is given by:
\[ DE + EF + FD = 4R \]

3. **Using the Lengths of the Feet of Altitudes**: The lengths of the segments formed by the feet of the altitudes can be expressed in terms of the sides of triangle ABC:
- \( DE = b \cos C \)
- \( EF = c \cos A \)
- \( FD = a \cos B \)

Thus, we can rewrite the perimeter equation as:
\[ a \cos B + b \cos C + c \cos A = 4R \]

4. **Applying the Sine Rule**: According to the sine rule, we have:
\[ a = 2R \sin A, \quad b = 2R \sin B, \quad c = 2R \sin C \]

5. **Substituting into the Perimeter Equation**: Substitute the expressions for a, b, and c into the perimeter equation:
\[ (2R \sin A \cos B) + (2R \sin B \cos C) + (2R \sin C \cos A) = 4R \]

6. **Simplifying the Equation**: Dividing through by \( 2R \) (assuming \( R \neq 0 \)):
\[ \sin A \cos B + \sin B \cos C + \sin C \cos A = 2 \]

7. **Using the Double Angle Identity**: Recall that \( \sin A \cos B = \frac{1}{2} \sin(2A) \). Thus, we can express the left-hand side as:
\[ \frac{1}{2} \sin(2A) + \frac{1}{2} \sin(2B) + \frac{1}{2} \sin(2C) = 2 \]
Multiplying through by 2 gives:
\[ \sin(2A) + \sin(2B) + \sin(2C) = 4 \]

8. **Analyzing the Sine Function**: The maximum value of \( \sin(x) \) is 1. Therefore, the maximum value of \( \sin(2A) + \sin(2B) + \sin(2C) \) can be at most 3. The equation \( \sin(2A) + \sin(2B) + \sin(2C) = 4 \) cannot hold true because it exceeds the maximum possible sum of 3.

9. **Conclusion**: Since the condition leads to a contradiction, the only way this can happen is if triangle ABC is a right-angled triangle. In a right triangle, one angle is \( 90^\circ \) (or \( \frac{\pi}{2} \)), which makes the sine of that angle equal to 1, thus allowing for the possibility of the equation to balance out.

### Final Answer:
Triangle ABC is a **right-angled triangle**.