To find the length of $AD$ in the given trapezium $ABCD$:

Step 1: Identify the key components.

• $AB$ is a vertical side (15 cm).
• $BC$ is the bottom horizontal side (30 cm).
• $CD$ is the slanted side (17 cm).
• $AD$ is the top horizontal side (to be determined).
• The trapezium has a perpendicular drawn from $D$ to $AB$, forming a right triangle $\triangle CDE$ with:
  • $DE$ as the height (equal to $AB = 15 \, \text{cm}$).
  • $CE$ as the base of the right triangle.

Step 2: Use the Pythagoras theorem to find $CE$.

In $\triangle CDE$: $CD^2 = CE^2 + DE^2$ Substituting values: $17^2 = CE^2 + 15^2$ $289 = CE^2 + 225$ $CE^2 = 289 - 225 = 64$ $CE = \sqrt{64} = 8 \, \text{cm}.$

Step 3: Determine $AD$.

Since $CE = 8 \, \text{cm}$, the remaining length along $BC$ is: $BE = BC - CE = 30 - 8 = 22 \, \text{cm}.$ Now, $AD$, the top horizontal side of the trapezium, equals $BE$: $AD = BE = 22 \, \text{cm}.$

Final Answer:

The length of $AD$ is 22 cm.

Do you want additional details or further explanation? Here are five related questions to consider:

1. How is the Pythagoras theorem applied in similar geometric problems?
2. How would you calculate the area of trapezium $ABCD$ using the given dimensions?
3. What if the height of the trapezium was different? How would that affect $AD$?
4. Can you generalize this solution for any trapezium with a slanted side?
5. What are some real-world applications of solving trapezium problems?

Tip: Always ensure the perpendicular dimensions are correctly identified to avoid errors in trapezium problems.