The image shows a right-angled triangle $\triangle ABD$ inside a rectangle $ABCD$, where $BC = 9 \, \text{m}$ and $CD = 20 \, \text{m}$. You are asked to find the area of $\triangle ABD$.

Step-by-step solution:

1. Find the area of the rectangle $ABCD$: The area of the rectangle is: $\text{Area of } ABCD = \text{length} \times \text{width} = BC \times CD = 9 \times 20 = 180 \, \text{m}^2$

2. Use the triangle area formula: The area of a triangle is given by: $\text{Area of } \triangle ABD = \frac{1}{2} \times \text{base} \times \text{height}$ In this case, the base is $AD = 20 \, \text{m}$ (since $AD$ is the same as the length of the rectangle), and the height is $AB = 9 \, \text{m}$ (since $AB$ is the width of the rectangle).

3. Calculate the area: $\text{Area of } \triangle ABD = \frac{1}{2} \times 20 \times 9 = \frac{1}{2} \times 180 = 90 \, \text{m}^2$

Thus, the area of $\triangle ABD$ is 90 m².

Would you like more details or have any questions about this process?

Here are some related questions for further understanding:

1. How do you calculate the hypotenuse of $\triangle ABC$ using the Pythagorean theorem?
2. What is the area of triangle $\triangle ABC$?
3. Can you calculate the area of triangle $\triangle BCD$?
4. How would the area of $\triangle ABD$ change if $AB$ was 10 meters instead of 9 meters?
5. How can the area of triangle $\triangle ABD$ be related to the area of the rectangle?

Tip: Always check if any triangle involved is a right-angled triangle, as it simplifies many calculations using standard geometric properties.