If S is any point on the side QR of a `DeltaPQR`, then

Options

A. `PQ+QR+RP gt 2PS`
B. `PQ+QR+RP lt 2PS`
C. `PQ+QR+RP=3PS`
D. None of these

Answer

To solve the problem, we need to analyze the triangle \( \Delta PQR \) and the point \( S \) located on the side \( QR \). We need to show that the sum of the lengths of the sides of the triangle is greater than twice the length of \( PS \).

### Step-by-Step Solution:

1. **Identify the Triangle and Points**:
We have triangle \( PQR \) with points \( P, Q, R \) and a point \( S \) on the side \( QR \).

2. **Draw the Triangle**:
Visualize or draw triangle \( PQR \) and mark point \( S \) on line segment \( QR \).

3. **Apply Triangle Inequality**:
According to the triangle inequality theorem, in any triangle, the sum of the lengths of any two sides must be greater than the length of the third side.

4. **Consider Triangle \( PQS \)**:
For triangle \( PQS \):
\[ PQ + QS > PS \quad \text{(Equation 1)} \]

5. **Consider Triangle \( PSR \)**:
For triangle \( PSR \):
\[ PR + SR > PS \quad \text{(Equation 2)} \]

6. **Add the Two Inequalities**:
Now, we add Equation 1 and Equation 2:
\[ (PQ + QS) + (PR + SR) > PS + PS \]
This simplifies to:
\[ PQ + QS + PR + SR > 2PS \]

7. **Recognize \( QS + SR = QR \)**:
Notice that \( QS + SR \) is equal to \( QR \). Therefore, we can rewrite the inequality as:
\[ PQ + QR + PR > 2PS \]

8. **Conclusion**:
We have shown that:
\[ PQ + QR + PR > 2PS \]
Therefore, the correct option is that \( PQ + QR + PR \) is greater than \( 2PS \).

### Final Answer:
The correct conclusion is that \( PQ + QR + PR > 2PS \).

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