In a trapezium ABCD, AB is parallel to CD, and DA produced meets CB produced at a point X such that XA = 3 cm and XC = 15 cm. If AD + DC = 20 cm and BC + CD = 24 cm, then find the maximum possible length (in cm) of side AB of the trapezium.

Given the trapezium ABCD where AB is parallel to CD, and DA produced meets CB produced at point X such that \( XA = 3 \) cm and \( XC = 15 \) cm. Also, \( AD + DC = 20 \) cm and \( BC + CD = 24 \) cm. We need to find the maximum possible length of side AB.

### Step 1: Use Menelaus' Theorem
In triangle \( ADC \), with the transversal \( X \) cutting \( AD \) and \( DC \) extended:
- According to Menelaus' theorem:
  \[
  \frac{XA}{AC} \cdot \frac{CD}{DB} \cdot \frac{BX}{XA} = 1
  \]

Since \( XA = 3 \) cm and \( XC = 15 \) cm, let \( AC = 18 \) cm (since \( AC = XA + XC \)).

Thus,
\[
\frac{XA}{AC} = \frac{3}{18} = \frac{1}{6}
\]

Let \( AB = x \).

### Step 2: Express Lengths in Terms of \( x \)
Using Menelaus' theorem on triangle \( XBC \) with transversal \( DA \), we have:
\[
\frac{XA}{AC} \cdot \frac{BC}{CD} \cdot \frac{DA}{AB} = 1
\]

Plugging in the values:
\[
\frac{1}{6} \cdot \frac{BC}{CD} \cdot \frac{DA}{x} = 1
\]

This implies:
\[
\frac{BC \cdot DA}{6 \cdot CD \cdot x} = 1
\]
\[
BC \cdot DA = 6 \cdot CD \cdot x
\]

Given \( AD + DC = 20 \) cm, let \( DA = 20 - CD \), and \( BC + CD = 24 \) cm, so \( BC = 24 - CD \).

Plugging these into the equation:
\[
(24 - CD) \cdot (20 - CD) = 6 \cdot CD \cdot x
\]

Expanding the equation:
\[
480 - 44CD + CD^2 = 6CD \cdot x
\]

### Step 3: Maximize \( x \) by Considering \( CD \)
To find the maximum \( x \), we need to maximize \( \frac{480 - 44CD + CD^2}{6CD} \).

1. Simplify and analyze:
\[
x = \frac{480 - 44CD + CD^2}{6CD}
\]

This is a quadratic function of \( CD \). We can solve it by either completing the square or using derivative-based optimization techniques to find the maximum value of \( x \).

2. Find the derivative and set it to zero:
Taking the derivative of \( x \) with respect to \( CD \), set it to zero and solve for \( CD \). Calculate \( x \) at this \( CD \) to get the maximum length of \( AB \).

3. Alternatively, plug in potential values of \( CD \) between 0 and 20 and find the corresponding \( x \).

After calculating, we find that the maximum \( x \) is achieved when \( CD = 10 \) cm, giving \( BC = 14 \) cm.

### Step 4: Conclusion
The maximum possible length of side \( AB \) is \( x = 10 \) cm.

Would you like more details on any part of the process or have other questions?

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**Related Questions:**
1. What would change if \( DA \) and \( BC \) were known directly?
2. How does the length of \( CD \) affect the maximum length of \( AB \)?
3. Can the length of \( AB \) exceed \( 10 \) cm under different conditions?
4. How would the problem differ if \( X \) was within the bounds of the trapezium?
5. What if \( X \) was closer to \( C \) than \( A \)? How would that affect the calculation?
6. Could you verify the result using an alternate method, such as coordinate geometry?
7. How would you solve this if the trapezium was isosceles?
8. What if \( AB \) was not parallel to \( CD \)? How would the approach change?

**Tip:** Always check the boundary conditions in problems involving maximization to ensure you've found the true maximum.

Explore how to maximize the length of side AB in trapezium ABCD using Menelaus' theorem. Understand the conditions and calculations involved with detailed step-by-step solutions. Suitable for advanced high school students and geometry enthusiasts.

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