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. Then
(i) The length of the other diagonal is
(ii) the length of side `AB` is equal to

Explanation

To solve the problem step by step, we will follow the information given and the properties of parallelograms.

### Step 1: Identify the equations of the diagonals
The equations of the diagonals are given as:
- Diagonal AC: \( x + 2y = 3 \)
- Diagonal BD: \( 2x + y = 3 \)

### Step 2: Find the slopes of the diagonals
To find the slopes, we can rewrite the equations in slope-intercept form \( y = mx + c \).

For diagonal AC:
\[ x + 2y = 3 \implies 2y = -x + 3 \implies y = -\frac{1}{2}x + \frac{3}{2} \]
Thus, the slope \( m_1 \) of diagonal AC is \( -\frac{1}{2} \).

For diagonal BD:
\[ 2x + y = 3 \implies y = -2x + 3 \]
Thus, the slope \( m_2 \) of diagonal BD is \( -2 \).

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

### Step 4: Find \( \sin \theta \)
Using the Pythagorean theorem, we can find \( \sin \theta \):
If \( \tan \theta = \frac{3}{4} \), then we can form a right triangle where the opposite side is 3 and the adjacent side is 4. The hypotenuse \( h \) is:
\[ h = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = 5 \]
Thus,
\[ \sin \theta = \frac{3}{5} \]

### Step 5: Use the area formula of the parallelogram
The area \( A \) of the parallelogram can be expressed in terms of the diagonals:
\[ A = \frac{1}{2} \times AC \times BD \times \sin \theta \]
We know:
- \( A = 8 \)
- \( AC = 4 \)
- \( \sin \theta = \frac{3}{5} \)

Substituting these values into the area formula:
\[ 8 = \frac{1}{2} \times 4 \times BD \times \frac{3}{5} \]
Simplifying:
\[ 8 = \frac{2}{5} \times BD \implies BD = 8 \times \frac{5}{2} = 20 \]

### Step 6: Find the length of side \( AB \)
Using the property of triangles, we can find the length of side \( AB \). The diagonals bisect each other, so:
- \( AO = OC = 2 \) (since \( AC = 4 \))
- \( BO = OD = 10/3 \) (since \( BD = 20/3 \))

Using the cosine rule:
\[ \cos \theta = \frac{AO^2 + OB^2 - AB^2}{2 \cdot AO \cdot OB} \]
We know \( \sin \theta = \frac{3}{5} \), hence \( \cos \theta = \sqrt{1 - \sin^2 \theta} = \frac{4}{5} \).

Substituting the values:
\[ \frac{4}{5} = \frac{2^2 + \left(\frac{10}{3}\right)^2 - AB^2}{2 \cdot 2 \cdot \frac{10}{3}} \]
Calculating:
\[ \frac{4}{5} = \frac{4 + \frac{100}{9} - AB^2}{\frac{40}{3}} \implies 4 \cdot \frac{40}{3} = 5 \left(4 + \frac{100}{9} - AB^2\right) \]
Simplifying:
\[ \frac{160}{3} = 20 + \frac{500}{9} - 5AB^2 \]
Clearing fractions and solving for \( AB^2 \):
\[ \frac{160 \cdot 9}{3} = 60 + 500 - 45AB^2 \]
\[ 480 = 60 + 500 - 45AB^2 \implies 45AB^2 = 60 + 500 - 480 \implies 45AB^2 = 80 \implies AB^2 = \frac{80}{45} = \frac{16}{9} \]
Thus,
\[ AB = \frac{4}{3} \]

### Final Answers
(i) The length of the other diagonal \( BD = \frac{20}{3} \) units.
(ii) The length of side \( AB = \frac{4}{3} \) units.