In a triangle ABC if `2Delta^(2)=(a^(2)b^(2)c^(2))/(a^(2)+b^(2)+c^(2))`, then it is

Options

A. equilateral
B. isosceles but not right angled
C. isosceles right angled
D. right angled

Explanation

To solve the problem, we start with the given equation:

\[ 2\Delta^2 = \frac{a^2b^2c^2}{a^2 + b^2 + c^2} \]

where \(\Delta\) is the area of triangle \(ABC\) with sides \(a\), \(b\), and \(c\).

### Step 1: Rearranging the Equation
We can cross-multiply to rearrange the equation:

\[ 2\Delta^2 (a^2 + b^2 + c^2) = a^2b^2c^2 \]

### Step 2: Using the Area Formula
We know that the area \(\Delta\) of triangle \(ABC\) can also be expressed using the circumradius \(R\):

\[ \Delta = \frac{abc}{4R} \]

Substituting this into our equation gives:

\[ 2\left(\frac{abc}{4R}\right)^2 (a^2 + b^2 + c^2) = a^2b^2c^2 \]

### Step 3: Simplifying the Equation
This simplifies to:

\[ 2 \cdot \frac{a^2b^2c^2}{16R^2} (a^2 + b^2 + c^2) = a^2b^2c^2 \]

Cancelling \(a^2b^2c^2\) from both sides (assuming \(a, b, c \neq 0\)) leads to:

\[ \frac{2(a^2 + b^2 + c^2)}{16R^2} = 1 \]

### Step 4: Further Simplification
This can be simplified to:

\[ a^2 + b^2 + c^2 = 8R^2 \]

### Step 5: Applying the Law of Cosines
Using the law of cosines, we know:

\[ a^2 + b^2 + c^2 = 2R^2 + 2R^2(\cos A + \cos B + \cos C) \]

We can substitute this into our equation:

\[ 2R^2 + 2R^2(\cos A + \cos B + \cos C) = 8R^2 \]

### Step 6: Solving for Cosines
This simplifies to:

\[ \cos A + \cos B + \cos C = 3 \]

### Step 7: Analyzing the Cosine Values
The maximum value of \(\cos A + \cos B + \cos C\) occurs when all angles are \(60^\circ\) (for an equilateral triangle), giving a maximum of \(3/2\). Thus, if \(\cos A + \cos B + \cos C = 3\), it implies that one of the angles must be \(90^\circ\).

### Conclusion
Thus, we conclude that triangle \(ABC\) must be a right-angled triangle.

### Final Answer
The triangle is a **right-angled triangle**.

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