If in `Delta ABC, 8R^(2) = a^(2) + b^(2) + c^(2)`, then the triangle ABC is

Options

A. right angled
B. isosceles
C. equilateral
D. none of these

Explanation

To solve the problem, we need to determine the type of triangle ABC given the equation \(8R^2 = a^2 + b^2 + c^2\).

### Step-by-Step Solution:

1. **Understanding the Given Equation**:
We start with the equation:
\[ 8R^2 = a^2 + b^2 + c^2 \]
Here, \(R\) is the circumradius of triangle ABC, and \(a\), \(b\), and \(c\) are the lengths of the sides opposite to angles \(A\), \(B\), and \(C\) respectively.

2. **Using the Relationship Between Sides and Angles**:
We know that the sides of the triangle can be expressed in terms of the circumradius \(R\) and the angles:
\[ a = 2R \sin A, \quad b = 2R \sin B, \quad c = 2R \sin C \]
Substituting these into the equation gives:
\[ a^2 + b^2 + c^2 = (2R \sin A)^2 + (2R \sin B)^2 + (2R \sin C)^2 \]
This simplifies to:
\[ a^2 + b^2 + c^2 = 4R^2 (\sin^2 A + \sin^2 B + \sin^2 C) \]

3. **Substituting Back into the Original Equation**:
Now, substituting this back into the original equation:
\[ 8R^2 = 4R^2 (\sin^2 A + \sin^2 B + \sin^2 C) \]
Dividing both sides by \(4R^2\) (assuming \(R \neq 0\)):
\[ 2 = \sin^2 A + \sin^2 B + \sin^2 C \]

4. **Using the Identity for Sine**:
We know that for any triangle:
\[ \sin^2 A + \sin^2 B + \sin^2 C \leq \frac{9}{4} \]
This maximum occurs when the triangle is equilateral. However, we have:
\[ \sin^2 A + \sin^2 B + \sin^2 C = 2 \]
This indicates that at least one of the angles must be \(90^\circ\) (since \(\sin^2 90^\circ = 1\) and the maximum sum of squares of sines for angles in a triangle is 2).

5. **Conclusion**:
Therefore, since the sum of the squares of the sines equals 2, it implies that one of the angles must be \(90^\circ\). Hence, triangle ABC is a right-angled triangle.

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