Let ABCD be a cyclic quadrilateral such that `AB=2, BC=3, angle B =120^(@)` and area of quadrilateral `=4sqrt3.` Which of the following is/are correct ?

Options

A. The value of `(AC)^(2)` is equal to 19
B. The sum of all positive value of product AC. BD I is equal to 35
C. The sum of all posible value of `(AD)^(2)` is equal to 29
D. The value of `(CD)^(2)` can be 4

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the Given Information
We have a cyclic quadrilateral \(ABCD\) with:
- \(AB = 2\)
- \(BC = 3\)
- \(\angle B = 120^\circ\)
- Area of quadrilateral \(ABCD = 4\sqrt{3}\)

### Step 2: Calculate the Area of Triangle \(ABC\)
The area of triangle \(ABC\) can be calculated using the formula:
\[ \text{Area} = \frac{1}{2} \times AB \times BC \times \sin(\angle B) \]
Substituting the values:
\[ \text{Area}_{ABC} = \frac{1}{2} \times 2 \times 3 \times \sin(120^\circ) \]
Since \(\sin(120^\circ) = \frac{\sqrt{3}}{2}\):
\[ \text{Area}_{ABC} = \frac{1}{2} \times 2 \times 3 \times \frac{\sqrt{3}}{2} = \frac{3\sqrt{3}}{2} \]

### Step 3: Calculate the Area of Triangle \(ADC\)
The total area of quadrilateral \(ABCD\) is given as \(4\sqrt{3}\). Therefore, the area of triangle \(ADC\) can be calculated as:
\[ \text{Area}_{ADC} = \text{Area}_{ABCD} - \text{Area}_{ABC} \]
Substituting the known values:
\[ \text{Area}_{ADC} = 4\sqrt{3} - \frac{3\sqrt{3}}{2} = \frac{8\sqrt{3}}{2} - \frac{3\sqrt{3}}{2} = \frac{5\sqrt{3}}{2} \]

### Step 4: Set Up the Area Equation for Triangle \(ADC\)
Let \(AD = x\) and \(CD = y\). The area of triangle \(ADC\) can also be expressed as:
\[ \text{Area}_{ADC} = \frac{1}{2} \times x \times y \times \sin(60^\circ) \]
Since \(\sin(60^\circ) = \frac{\sqrt{3}}{2}\):
\[ \frac{5\sqrt{3}}{2} = \frac{1}{2} \times x \times y \times \frac{\sqrt{3}}{2} \]
Multiplying both sides by 2:
\[ 5\sqrt{3} = \frac{xy\sqrt{3}}{2} \]
Dividing by \(\sqrt{3}\):
\[ 5 = \frac{xy}{2} \implies xy = 10 \]

### Step 5: Use the Cosine Law in Triangle \(ABC\)
Using the cosine law, we can find \(AC^2\):
\[ AC^2 = AB^2 + BC^2 - 2 \times AB \times BC \times \cos(120^\circ) \]
Substituting the known values:
\[ AC^2 = 2^2 + 3^2 - 2 \times 2 \times 3 \times \left(-\frac{1}{2}\right) \]
Calculating:
\[ AC^2 = 4 + 9 + 6 = 19 \]

### Step 6: Set Up the Equation for \(AC^2\) Using \(x\) and \(y\)
Using the cosine law again for triangle \(ADC\):
\[ AC^2 = x^2 + y^2 - 2xy \cos(60^\circ) \]
Since \(\cos(60^\circ) = \frac{1}{2}\):
\[ 19 = x^2 + y^2 - xy \]

### Step 7: Substitute \(xy\) into the Equation
We know \(xy = 10\):
\[ 19 = x^2 + y^2 - 10 \implies x^2 + y^2 = 29 \]

### Step 8: Solve for \(x^2\) and \(y^2\)
Using the identity \((x+y)^2 = x^2 + y^2 + 2xy\):
\[ (x+y)^2 = 29 + 20 = 49 \implies x+y = 7 \]
Now, we can find \(x^2\) and \(y^2\):
\[ x^2 + y^2 = 29 \]
Let \(x^2 = a\) and \(y^2 = b\):
\[ a + b = 29 \quad \text{and} \quad ab = 100 \]
The roots of the quadratic equation \(t^2 - 7t + 10 = 0\) give us the values of \(x\) and \(y\).

### Step 9: Find Possible Values of \(AD^2\) and \(CD^2\)
The possible values of \(AD^2\) and \(CD^2\) are \(25\) and \(4\). Thus, the sum of all possible values of \(AD^2\) is:
\[ 25 + 4 = 29 \]

### Conclusion
The correct options based on the calculations are:
- \(AC^2 = 19\)
- \(AD^2 + CD^2 = 29\)