If `Delta` represents the area of acute angled triangle ABC `sqrt(a^(2)b^(2) -4Delta^(2)) + sqrt(b^(2) c^(2) -4Delta^(2)) + sqrt(c^(2) a^(2) -4Delta^(2))=`

Options

A. `a^(2)+ b^(2) + c^(2)`
B. `(a^(2) + b^(2) + c^(2))/(2)`
C. `ab cos C + bc cos A + ca cos B`
D. `ab sin C + bc sin A + ca sin B`

Explanation

To solve the problem, we need to evaluate the expression:

\[ \sqrt{a^2b^2 - 4\Delta^2} + \sqrt{b^2c^2 - 4\Delta^2} + \sqrt{c^2a^2 - 4\Delta^2} \]

where \(\Delta\) is the area of triangle \(ABC\).

### Step 1: Use the formula for the area of a triangle
We know that the area \(\Delta\) of triangle \(ABC\) can be expressed in terms of its sides \(a\), \(b\), and \(c\) as:

\[ \Delta = \frac{1}{2}ab \sin C = \frac{1}{2}bc \sin A = \frac{1}{2}ca \sin B \]

### Step 2: Substitute \(\Delta\) into the expression
We can express \(4\Delta^2\) as follows:

\[ 4\Delta^2 = 4\left(\frac{1}{2}ab \sin C\right)^2 = a^2b^2 \sin^2 C \]

Thus, we can rewrite \(a^2b^2 - 4\Delta^2\):

\[ a^2b^2 - 4\Delta^2 = a^2b^2 - a^2b^2 \sin^2 C = a^2b^2(1 - \sin^2 C) = a^2b^2 \cos^2 C \]

### Step 3: Apply the same for the other terms
Similarly, we can derive the other two terms:

\[ b^2c^2 - 4\Delta^2 = b^2c^2 \cos^2 A \]
\[ c^2a^2 - 4\Delta^2 = c^2a^2 \cos^2 B \]

### Step 4: Substitute back into the original expression
Now substituting these results back into the original expression:

\[ \sqrt{a^2b^2 \cos^2 C} + \sqrt{b^2c^2 \cos^2 A} + \sqrt{c^2a^2 \cos^2 B} \]

This simplifies to:

\[ ab \cos C + bc \cos A + ca \cos B \]

### Step 5: Use the cosine rule
Using the cosine rule, we know that:

\[ ab \cos C + bc \cos A + ca \cos B = \frac{1}{2}(a^2 + b^2 + c^2) \]

### Final Result
Thus, we conclude that:

\[ \sqrt{a^2b^2 - 4\Delta^2} + \sqrt{b^2c^2 - 4\Delta^2} + \sqrt{c^2a^2 - 4\Delta^2} = \frac{1}{2}(a^2 + b^2 + c^2) \]

### Conclusion
The correct answer is:

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