Consider the following statements
I. Let ABCD be a parallelogram which is not a rectangle.
Then,`2(AB^2 +BC^2)** AC^2 + BD^2` II. If ABCD is a rhombus with AB = 4cm, then AC + BD = n for some positive integer n.
Which of the above statements is/are correct?

Options

A. a. Only I
B. b. Only II
C. c. Both I and II
D. d. Neither I nor II

Answer

To solve the problem, we need to analyze the two statements given about the parallelogram ABCD.

### Step 1: Analyze Statement I
**Statement I:** "Let ABCD be a parallelogram which is not a rectangle. Then, \(2(AB^2 + BC^2) \neq AC^2 + BD^2\)."

1. **Understanding the properties of a parallelogram:**
In any parallelogram, the diagonals bisect each other. The diagonals \(AC\) and \(BD\) can be analyzed using the properties of triangles formed by the sides and diagonals.

2. **Using the triangle properties:**
Consider triangles \(ABC\) and \(BCD\):
- In triangle \(ABC\), by the law of cosines, we have:
\[ AC^2 = AB^2 + BC^2 - 2 \cdot AB \cdot BC \cdot \cos(\angle ABC) \]
- In triangle \(BCD\), we have:
\[ BD^2 = BC^2 + CD^2 - 2 \cdot BC \cdot CD \cdot \cos(\angle BCD) \]
- Since \(AB = CD\) in a parallelogram, we can substitute \(CD\) with \(AB\).

3. **Combining the equations:**
Adding both equations:
\[ AC^2 + BD^2 = (AB^2 + BC^2 - 2 \cdot AB \cdot BC \cdot \cos(\angle ABC)) + (BC^2 + AB^2 - 2 \cdot BC \cdot AB \cdot \cos(\angle BCD)) \]
This simplifies to:
\[ AC^2 + BD^2 = 2AB^2 + 2BC^2 - 2(AB \cdot BC)(\cos(\angle ABC) + \cos(\angle BCD)) \]
Since \(\angle ABC\) and \(\angle BCD\) are not equal in a non-rectangular parallelogram, we conclude that:
\[ 2(AB^2 + BC^2) \neq AC^2 + BD^2 \]
Thus, Statement I is correct.

### Step 2: Analyze Statement II
**Statement II:** "If ABCD is a rhombus with \(AB = 4 \text{ cm}\), then \(AC + BD = n\) for some positive integer \(n\)."

1. **Understanding the properties of a rhombus:**
In a rhombus, all sides are equal, and the diagonals bisect each other at right angles.

2. **Calculating the diagonals:**
Let \(AC\) and \(BD\) be the diagonals. Since \(AB = 4 \text{ cm}\), we can denote the half-lengths of the diagonals as:
\[ AO = \frac{AC}{2}, \quad BO = \frac{BD}{2} \]
By the Pythagorean theorem in triangle \(AOB\):
\[ AB^2 = AO^2 + BO^2 \]
Substituting \(AB = 4\):
\[ 4^2 = \left(\frac{AC}{2}\right)^2 + \left(\frac{BD}{2}\right)^2 \]
This simplifies to:
\[ 16 = \frac{AC^2}{4} + \frac{BD^2}{4} \]
Multiplying through by 4:
\[ 64 = AC^2 + BD^2 \]

3. **Finding \(AC + BD\):**
The values of \(AC\) and \(BD\) can be any positive integers that satisfy the equation \(AC^2 + BD^2 = 64\). For example, if \(AC = 8\) and \(BD = 0\), or \(AC = 4\sqrt{2}\) and \(BD = 4\sqrt{2}\), etc. Hence, \(AC + BD\) can be expressed as a positive integer \(n\).

Thus, Statement II is also correct.

### Conclusion:
- **Statement I:** Correct
- **Statement II:** Correct

### Final Answer:
Both statements are correct.

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