`DeltaABC` has different side lengths a,b,c. If `a^(2),b^(2),c^(2)` as sides form another `DeltaPQR`, then `DeltaABC` will always be

Options

A. acute angled triangle only
B. obtuse angled triangle only
C. sometimes acute or sometimes obtuse depending on values of a,b nd c
D. none of these

Explanation

To determine the nature of triangle ABC given that the sides of triangle PQR are \(a^2\), \(b^2\), and \(c^2\), we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Problem**:
We have triangle ABC with sides \(a\), \(b\), and \(c\). We need to analyze triangle PQR formed by the squares of these sides: \(a^2\), \(b^2\), and \(c^2\).

2. **Triangle Inequality**:
For any triangle, the sum of the lengths of any two sides must be greater than the length of the third side. Thus, for triangle ABC, we have:
\[ a + b > c, \quad b + c > a, \quad c + a > b \]
For triangle PQR, we need to check:
\[ a^2 + b^2 > c^2, \quad b^2 + c^2 > a^2, \quad c^2 + a^2 > b^2 \]

3. **Applying the Triangle Inequality to Triangle PQR**:
Using the triangle inequality on triangle PQR:
- From \(a + b > c\), squaring both sides gives:
\[ (a + b)^2 > c^2 \implies a^2 + 2ab + b^2 > c^2 \]
- This shows that \(a^2 + b^2 > c^2\) is satisfied if \(2ab > 0\), which is true since \(a\) and \(b\) are positive.

- Similarly, for the other inequalities:
\[ b^2 + c^2 > a^2 \quad \text{and} \quad c^2 + a^2 > b^2 \]
can be shown to hold true using the same reasoning.

4. **Finding the Angles**:
To determine the type of triangle ABC, we can use the cosine rule:
\[ \cos A = \frac{b^2 + c^2 - a^2}{2bc} \]
Since \(b^2 + c^2 > a^2\) (as shown above), we have:
\[ \cos A > 0 \implies A < 90^\circ \]
By similar calculations, we can show that:
\[ \cos B > 0 \quad \text{and} \quad \cos C > 0 \]
Therefore, angles \(B\) and \(C\) are also less than \(90^\circ\).

5. **Conclusion**:
Since all angles \(A\), \(B\), and \(C\) in triangle ABC are less than \(90^\circ\), triangle ABC is an **acute-angled triangle**.

### Final Answer:
Triangle ABC will always be an **acute-angled triangle**.