□ABCD is a trapezium where AB ∥ CD. It is given that ℓ(AB) = 14 cm and ℓ(CD) = 22 cm. If the two diagonals AC and BD intersect at point O and the area of the trapezium is 324 sq. cm then the area (in sq. cm) of ∆AOD will be ___________ .

To solve this problem, let's use the properties of trapeziums and triangles. Given that \( AB \parallel CD \) in trapezium \( ABCD \), we can apply the concept of similar triangles and the area of the trapezium.

### Step 1: Determine the height of the trapezium
The area of a trapezium is given by:

\[
\text{Area} = \frac{1}{2} \times (\text{Base}_1 + \text{Base}_2) \times \text{Height}
\]

Here, \( \text{Base}_1 = AB = 14 \) cm, \( \text{Base}_2 = CD = 22 \) cm, and the area is given as \( 324 \) sq. cm.

Let's denote the height of the trapezium by \( h \). Then,

\[
324 = \frac{1}{2} \times (14 + 22) \times h
\]

\[
324 = \frac{1}{2} \times 36 \times h
\]

\[
324 = 18h
\]

\[
h = \frac{324}{18} = 18 \text{ cm}
\]

### Step 2: Apply the area ratio in similar triangles
Since \( AB \parallel CD \), triangles \( \Delta AOD \) and \( \Delta BOC \) are similar. The ratio of their areas will be the square of the ratio of corresponding sides.

The diagonals of the trapezium divide it into four triangles: \( \Delta AOD \), \( \Delta BOC \), \( \Delta ABO \), and \( \Delta COD \).

### Step 3: Use the formula for area based on similar triangles
Since \( AB \parallel CD \), triangles \( \Delta AOD \) and \( \Delta BOC \) are similar, and the ratio of their areas will be the square of the ratio of the corresponding sides. The sides \( AB \) and \( CD \) are corresponding sides in the similar triangles, so:

\[
\frac{\text{Area of } \Delta AOD}{\text{Area of } \Delta BOC} = \left(\frac{AB}{CD}\right)^2 = \left(\frac{14}{22}\right)^2 = \left(\frac{7}{11}\right)^2 = \frac{49}{121}
\]

Let the area of \( \Delta AOD \) be \( x \) and the area of \( \Delta BOC \) be \( y \). Then,

\[
\frac{x}{y} = \frac{49}{121}
\]

Given that the total area of the trapezium is the sum of the areas of the four triangles, and \( \text{Area of } \Delta AOD + \text{Area of } \Delta BOC = \text{Area of } \Delta ABO + \text{Area of } \Delta COD \).

We also know that:

\[
\text{Area of } \Delta AOD + \text{Area of } \Delta BOC = \frac{49x + 121x}{121} = 324 \text{ cm}^2
\]

Let the area of \( \Delta AOD \) be \( x \). Then:

\[
x + y = 324 \text{ cm}^2
\]

\(\Rightarrow x = \frac{49}{170} \times 324 \)

### Step 4: Calculate the area of \( \Delta AOD \)
Finally, we solve for \( x \) (Area of \( \Delta AOD \)):

\[
x = \frac{49}{170} \times 324
\]

\[
x = \frac{49 \times 324}{170} \approx 93.2 \text{ sq. cm}
\]

Therefore, the area of \( \Delta AOD \) is approximately \( 93 \) sq. cm.

### Conclusion:
The area of \( \Delta AOD \) is **93 sq. cm**.

Would you like to go into more details or have any other questions? Here are some related questions you might be interested in:

1. How does the concept of similar triangles apply to trapeziums?
2. Can we solve this problem using other geometric properties?
3. What is the significance of the diagonals intersecting in trapeziums?
4. How do we calculate the area of other quadrilaterals?
5. What are the key properties of trapeziums?

**Tip:** Always remember to double-check whether the problem involves similar triangles or other relationships that can simplify area calculations.

Learn how to calculate the area of triangle AOD in trapezium ABCD where AB is parallel to CD, AB = 14 cm, CD = 22 cm, and the total area of the trapezium is 324 sq. cm. Understand the application of similar triangles and the formula for area in trapeziums.

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