Consider `P,Q,R` to be vertices with integral coordinates and `(|PR|+|RQ|)^(2)lt8`. Area `(/_\PQR)+1` then

Options

A. `/_R` can be a right angle
B. `/_\PQR`can be isosceles
C. `P,Q,R` can lie on a square
D. `P,Q,R` can lie on circle centred on midpoint of line segment `PQ`

Explanation

To solve the problem, we need to analyze the given condition and derive the necessary conclusions about the triangle PQR.

### Step-by-Step Solution:

1. **Understanding the Given Condition**:
We are given that \( |PR| + |RQ| < \sqrt{8} \). This implies that the sum of the lengths of segments PR and RQ must be less than \( \sqrt{8} \).

2. **Using the Triangle Inequality**:
According to the triangle inequality, the length of any side of a triangle must be less than the sum of the lengths of the other two sides. Therefore, we can say:
\[ |PQ| < |PR| + |RQ| \]
This means that \( |PQ| \) must also be less than \( \sqrt{8} \).

3. **Area of Triangle PQR**:
The area of triangle PQR can be expressed in terms of its vertices. For vertices with integral coordinates, the area can be calculated using the formula:
\[ \text{Area} = \frac{1}{2} \times |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)| \]
However, we will focus on the implications of the given conditions rather than calculating the area explicitly.

4. **Inequality Analysis**:
Since \( |PR| + |RQ| < \sqrt{8} \), we can square both sides:
\[ (|PR| + |RQ|)^2 < 8 \]
Expanding this gives:
\[ |PR|^2 + |RQ|^2 + 2|PR||RQ| < 8 \]

5. **Relating Area to Lengths**:
We know that the area of triangle PQR can also be related to the lengths of its sides. The area \( A \) can be expressed in terms of the lengths \( |PR| \) and \( |RQ| \):
\[ A = \frac{1}{2} \times |PR| \times |RQ| \times \sin(\theta) \]
where \( \theta \) is the angle between sides PR and RQ.

6. **Conclusion about Triangle Types**:
From the inequalities derived, we can conclude:
- If \( |PR| + |RQ| < \sqrt{8} \), then the triangle can be a right triangle at vertex R.
- The triangle can also be isosceles if \( |PR| = |RQ| \).
- The triangle can lie within a square or on a circle centered at the midpoint of line segment PQ.

### Final Answer:
All options given in the problem are correct:
- A: Triangle PQR can have a right angle at R.
- B: Triangle PQR can be isosceles.
- C: Triangle PQR can lie on a square.
- D: Triangle PQR can lie on a circle centered on the midpoint of line segment PQ.