Let ABCD be a parallelogram the equation of whose diagonals are `AC : x+2y =3`; BD: 2x + y = 3. If length of diagonal` AC =4` units and area of `ABCD = 8` sq. units. Find the length of the other diagonal is a. `10//3` b. `2` c. `20//3` d. None of these

Options

A. `10//3`
B. 2
C. `20//3`
D. none of these

Explanation

To find the length of the other diagonal \( BD \) of the parallelogram \( ABCD \), we will follow these steps:

### Step 1: Understand the properties of the parallelogram
In a parallelogram, the diagonals bisect each other. This means that if \( O \) is the intersection point of the diagonals \( AC \) and \( BD \), then:
- \( OA = OC \)
- \( OB = OD \)

### Step 2: Use the area of the parallelogram
The area of parallelogram \( ABCD \) can be expressed in terms of the lengths of the diagonals and the sine of the angle \( \theta \) between them:
\[ \text{Area} = \frac{1}{2} \times AC \times BD \times \sin(\theta) \]
Given that the area is \( 8 \) square units and the length of diagonal \( AC \) is \( 4 \) units, we can substitute these values into the equation:
\[ 8 = \frac{1}{2} \times 4 \times BD \times \sin(\theta) \]
This simplifies to:
\[ 8 = 2 \times BD \times \sin(\theta) \]
\[ BD \times \sin(\theta) = 4 \quad \text{(Equation 1)} \]

### Step 3: Find the slopes of the diagonals
The equations of the diagonals are given as:
- \( AC: x + 2y = 3 \)
- \( BD: 2x + y = 3 \)

To find the slopes, we can rewrite these equations in slope-intercept form \( y = mx + b \).

For \( AC \):
\[ 2y = -x + 3 \implies y = -\frac{1}{2}x + \frac{3}{2} \quad \Rightarrow \text{slope } m_1 = -\frac{1}{2} \]

For \( BD \):
\[ y = -2x + 3 \quad \Rightarrow \text{slope } m_2 = -2 \]

### Step 4: Calculate the angle between the diagonals
The angle \( \theta \) between the two lines can be found using the formula:
\[ \tan(\theta) = \frac{m_2 - m_1}{1 + m_1 m_2} \]
Substituting the slopes:
\[ \tan(\theta) = \frac{-2 - (-\frac{1}{2})}{1 + (-\frac{1}{2})(-2)} = \frac{-2 + \frac{1}{2}}{1 + 1} = \frac{-\frac{4}{2} + \frac{1}{2}}{2} = \frac{-\frac{3}{2}}{2} = -\frac{3}{4} \]

### Step 5: Find \( \sin(\theta) \)
Using the identity \( \sin^2(\theta) + \cos^2(\theta) = 1 \):
Let \( \sin(\theta) = \frac{3}{5} \) and \( \cos(\theta) = \frac{4}{5} \) (from the right triangle formed).

### Step 6: Substitute \( \sin(\theta) \) back into Equation 1
Now we can substitute \( \sin(\theta) \) back into Equation 1:
\[ BD \times \frac{3}{5} = 4 \]
Solving for \( BD \):
\[ BD = 4 \times \frac{5}{3} = \frac{20}{3} \]

### Conclusion
Thus, the length of the other diagonal \( BD \) is \( \frac{20}{3} \).